Class 12 - Maths - Integrals

Exercise 7.1

Find an anti-derivative (or integral) of the following functions by the method of inspection.

Question 1:

sin 2x

The anti-derivative of sin 2x is a function of x whose derivative is sin 2x.

It is known that,

d(cos 2x)/dx = -2 sin 2x

=> sin 2x = (-1/2) * d(cos 2x)/dx

=> sin 2x = d[(-1/2) * cos 2x]/dx

Therefore, the anti-derivative of sin 2x is (-1/2) * cos 2x

Question 2:

cos 3x

The anti-derivative of cos 3x is a function of x whose derivative is cos 3x.

It is known that,

d(sin 3x)/dx = 3 cos 3x

=> cos 3x = (1/3) * d(sin 3x)/dx

=> cos 3x = d[(1/3) * cos 2x]/dx

Therefore, the anti-derivative of cos 3x is (1/3) * sin 3x

Question 3:

e2x

The anti-derivative of e2x is the function of x whose derivative is e2x

It is known that,

d(e2x)/dx = 2e2x

=> e2x = (1/2) * d(e2x)/dx

=> e2x = d[(1/2) * e2x]/dx

Therefore, the anti-derivative of e2x is (1/2) * e2x

Question 4:

(ax + b)2

The anti-derivative of (ax + b)2 is the function of x whose derivative is (ax + b)2

It is known that,

d[(ax + b)3]/dx = 3a(ax + b)2

=> (ax + b)2 = (1/3a) * d[(ax + b)3]/dx

=> (ax + b)2 = d[(1/3a) * (ax + b)3]/dx

Therefore, the anti-derivative of (ax + b)2 is (1/3a) * (ax + b)3

Question 5:

sin 2x – 4e3x

The anti-derivative of sin 2x – 4e3x is the function of x whose derivative is sin2x – 4e3x

It is known that,

d[(1-/2) * cos 2x – (4/3) * e3x]/dx = sin2x – 4e3x

Therefore, the anti derivative of sin2x – 4e3x is (1-/2) * cos 2x – (4/3) * e3x

Find the following integrals in Exercises 6 to 20:

Question 6:

ʃ (4e3x + 1) dx

Given, ʃ (4e3x + 1) dx = ʃ 4e3x dx + ʃ dx

= 4ʃ e3x dx + ʃ dx

= (4/3) * e3x + x + C

So,  ʃ (4e3x + 1) dx = (4/3) * e3x + x + C

Question 7:

ʃ [x2 * (1 – 1/x2)] dx

Given, ʃ[x2 * (1 – 1/x2)] dx = ʃ (x2 – 1) dx

= ʃ x2 dx – ʃ dx

= x2/3 – x + C

So, ʃ [x2 * (1 – 1/x2)] dx = x2/3 – x + C

Question 8:

ʃ (ax2 + bx + c) dx

ʃ (ax2 + bx + c) dx = ʃ ax2 dx + ʃ bx dx + ʃ c dx

= aʃ x2 dx + bʃ x dx + c ʃ dx

= ax3/3 + bx2/2 dx + cx + C

So, ʃ (ax2 + bx + c) dx = ax3/3 + bx2/2 dx + cx + C

Question 9:

ʃ (2x2 + ex) dx

ʃ (2x2 + ex)dx = ʃ 2x2 dx + ʃ ex dx

= 2ʃ x2 dx + ʃ ex dx

= 2x3/3 + ex + C

So, ʃ (2x2 + ex)dx = 2x3/3 + ex + C

Question 10:

ʃ (√x – 1/√x)2 dx

ʃ (√x – 1/√x)2 dx = ʃ (x + 1/x - 2) dx

= ʃx dx + ʃ(1/x) dx - 2ʃdx

= x2/2 + log|x| - 2x + C

So, ʃ (√x – 1/√x)2 dx = x2/2 + log|x| - 2x + C

Question 11:

ʃ [(x3 + 5x2 - 4)/x2] dx

ʃ [(x3 + 5x2 - 4)/x2] dx = ʃ [(x + 5 - 4x-2] dx

= ʃx dx + 5ʃ dx - 4ʃ x-2 dx

= x2/2 + 5x – 4(x-1)/(-1) + C

= x2/2 + 5x + 4/x + C

So, ʃ [(x3 + 5x2 - 4)/x2] dx = x2/2 + 5x + 4/x + C

Question 12:

ʃ [(x3 + 3x + 4)/√x] dx

ʃ [(x3 + 3x + 4)/√x] dx = ʃ [(x3 + 3x + 4)/x1/2] dx

= ʃ [(x3-1/2 + 3x1 – 1/2 + 4x-1/2] dx

= ʃ [(x5/2 + 3x1/2 + 4x-1/2] dx

= ʃ [(x5/2 dx + ʃ 3x1/2 dx + ʃ 4x-1/2 dx

= ʃ [(x5/2 dx + 3 ʃ x1/2 dx + 4 ʃ x-1/2 dx

= x7/2/(7/2) + 3x3/2/(3/2) + 4x1/2/(1/2) + C

= 2x7/2/7 + 2x3/2 + 8√x + C

So, ʃ [(x3 + 3x + 4)/√x] dx = 2x7/2/7 + 2x3/2 + 8√x + C

Question 13:

ʃ [(x3 – x2 + x - 1)/(x - 1)] dx

ʃ [(x3 – x2 + x - 1)/(x - 1)] dx = ʃ (x2 + 1) dx             [on dividing]

= ʃ x2 dx + ʃ dx

= x3/3 + x + C

So, ʃ [(x3 – x2 + x - 1)/(x - 1)] dx = x3/3 + x + C

Question 14:

ʃ (1 - x)√x dx

ʃ (1 - x)√x dx = ʃ (√x – x3/2) dx

= ʃ x1/2 dx – ʃ x3/2 dx

= x3/2/(3/2) – x5/2/(5/2) + C

= 2x3/2/3 – 2x5/2/5 + C

So, ʃ (1 - x)√x dx = 2x3/2/3 – 2x5/2/5 + C

Question 15:

ʃ √x(3x2 + 2x + 3) dx

ʃ √x(3x2 + 2x + 3) dx = ʃ (3x5/2 + 2x3/2 + 3x1/2) dx

= 3ʃ x5/2 dx + 2ʃ x3/2 dx + 3ʃ x1/2 dx

= 3x7/2/(7/2) + 2x5/2/(5/2) + 3x3/2/(3/2) + C

= 6x7/2/7 + 4x5/2/5 + 2x3/2 + C

So, ʃ √x(3x2 + 2x + 3) dx = 6x7/2/7 + 4x5/2/5 + 2x3/2 + C

Question 16:

ʃ (2x – 3 cos x + ex) dx

ʃ (2x – 3 cos x + ex) dx = 2ʃ x dx – 3ʃ cos x dx + ʃ ex dx

= 2x2/2 – 3 sin x + ex + C

= x2 – 3 sin x + ex + C

So, ʃ (2x – 3 cos x + ex) dx = x2 – 3 sin x + ex + C

Question 17:

ʃ (2x2 – 3 sin x + 5√x) dx

ʃ (2x2 – 3 sin x + 5√x) dx = 2ʃ x2 dx – 3ʃ sin x dx + 5ʃ x1/2 dx

= 2x3/3 – 3(-cos x) + 5x3/2/(3/2) + C

= 2x3/3 + 3 cos x + 10x3/2/3 + C

So, ʃ (2x2 – 3 sin x + 5√x) dx = 2x3/3 + 3 cos x + 10x3/2/3 + C

Question 18:

ʃ sec x(sec x + tan x) dx

ʃ sec x(sec x + tan x) dx = ʃ (sec2 x + sec x * tan x) dx

= ʃ (sec2 x dx + ʃ (sec x * tan x) dx

= tan x + sec x + C

So, ʃ sec x(sec x + tan x) dx = tan x + sec x + C

Question 19:

ʃ sec2 x/cosec2 x dx

ʃ sec2 x/cosec2 x dx = ʃ [(1/cos2 x)/(1/sin2 x)] dx

= ʃ (sin2 x /cos2 x) dx

= ʃ tan2 x dx

= ʃ (sec2 x – 1) dx

= ʃ sec2 x dx – ʃ dx

= tan x – x + C

So, ʃ sec2 x/cosec2 x dx = tan x – x + C

Question 20:

ʃ (2 – 3 sin x)/cos2 x dx

ʃ (2 – 3 sin x)/cos2 x dx = ʃ (2/cos2 x – 3 sin x / cos2 x) dx

= ʃ (2 sec2 x dx – 3ʃ tan x * sec x dx

= 2 tan x – 3 sec x + C

So, ʃ (2 – 3 sin x)/cos2 x dx = 2 tan x – 3 sec x + C

Question 21:

The anti derivative of (√x + 1/√x) equals

(A) x1/3/3 + 2x1/2 + C                                                     (B) 2x2/3/3 + x2/2 + C

(C) 2x3/2/3 + 2x1/2 + C                                                   (D) 3x3/2/2 + x1/2/2 + C

ʃ (√x + 1/√x) dx = ʃ (x1/2 + x-1/2) dx

= ʃ x1/2 dx + ʃ x-1/2 dx

= x3/2/(3/2) + x1/2/(1/2) + C

= 2x3/2/3 + 2x1/2 + C

Hence, the correct answer is option C.

Question 22:

If d[f(x)]/dx = 4x3 – 3/x4, such that f(2) = 0, then f(x) is

(A) x4 + 1/x3 – 129/8                                              (B) x3 + 1/x4 – 129/8

(C) x4 + 1/x3 + 129/8                                              (D) x3 + 1/x4 – 129/8

Given that

d[f(x)]/dx = 4x3 – 3/x4

So, anti derivative of 4x3 – 3/x4 = f(x)

Now, f(x) = ʃ (4x3 – 3/x4) dx

=> f(x) = ʃ 4x3 dx – ʃ (3/x4) dx

=> f(x) = 4ʃ x3 dx – 3ʃ (1/x4) dx

=> f(x) = 4ʃ x3 dx – 3ʃ x-4 dx

=> f(x) = 4x4/4 – 3x-3/(-3) + C

=> f(x) = x4 + 1/x3 + C

Also, f(2) = 0

=> 24 + 1/23 + C = 0

=> 16 + 1/8 + C = 0

=> 129/8 + C = 0

=> C = -129/8

So, f(x) = x4 + 1/x3 – 129/8

Hence, the correct answer is option A.

Exercise 7.2

Integrate the functions in Exercises 1 to 37:

Question 1:

2x/(1 + x2)

Let 1 + x2 = t

=> 2x dx = dt

Now, ʃ 2x/(1 + x2) dx = ʃ dt/t

= log|t| + C

= log|1 + x2| + C

= log(1 + x2) + C

Question 2:

(log x)2/x

Let log x = t

=> dx/x = dt

Now, ʃ [(log x)2/x] dx = ʃ t2 dt

= t3/3 + C

= (log|x|)3/3 + C

Question 3:

1/(x + x * log x)

1/(x + x * log x) = 1/{x(1 + log x)}

Let 1 + log x = t

=> dx/x = dt

Now, ʃdx/(x + x * log x) = ʃdt/t

= log|t| + C

= log|1 + log x| + C

Question 4:

sin x * sin(cos x)

Let cos x = t

=> -sin x dx = dt

Now, ʃ sin x * sin(cos x) = -ʃ sin t dt

= -[-cos t] + C

= cos t + C

= cos(cos x) + C

Question 5:

sin(ax + b) * cos(ax + b)

sin(ax + b) * cos(ax + b) = [2 * sin(ax + b) * cos(ax + b)]2

= sin 2(ax + b) /2                                     [Since sin 2x = 2 * sin x * cos x]

Let 2(ax + b) = t

=> 2a dx = dt

=> dx = dt/2a

Now, ʃ [sin 2(ax + b) /2]dx = (1/2) * ʃ (sin t /2a) dt

= (1/4a) * ʃ sin t dt

= (1/4a) * (-cos t) + C

= (-1/4a) * cos 2(ax + b) + C

Question 6:

√(ax + b)

Let ax + b = t

=> a dx = dt

=> dx = dt/a

Now, ʃ √(ax + b) dx = (1/a)ʃ √t dt

= (1/a)ʃ t1/2 dt

= (1/a) t3/2 (3/2) + C

= (2/3a) t3/2 + C

= (2/3a)(ax + b)3/2 + C

Question 7:

x√(x + 2)

Let x + 2 = t

=> dx = dt

Now, ʃ x√(x + 2) dx = ʃ (t - 2)√t dt

= ʃ (t - 2)t1/2 dt

= ʃ (t3/2 - 2t1/2) dt

= ʃ t3/2 dt - 2ʃ t1/2 dt

= t5/2/(5/2) – 2 * t3/2/(3/2) + C

= 2t5/2/5 – 4t3/2/3 + C

= 2(x + 2)5/2/5 – 4(x + 2)3/2/3 + C

Question 8:

x√(1 + 2x2)

Let 1 + 2x2 = t

=> 4x dx = dt

Now, ʃ x√(1 + 2x2) dx = ʃ (√t/4) dt

= (1/4)ʃ t1/2 dt

= (1/4) t3/2/(3/2) + C

= (1/6) t3/2 + C

= (1 + 2x)3/2/6 + C

Question 9:

(4x + 2)√(x2 + x + 1)

Let x2 + x + 1 = t

(2x + 1)dx = dt

Now, ʃ(4x + 2)√(x2 + x + 1) = ʃ2(2x + 1)√(x2 + x + 1)

= ʃ2√t dt

= 2ʃ t1/2 dt

= 2t3/2/(3/2) + C

= 4t3/2/3 + C

= 4(x2 +x + 1)3/2/3 + C

Question 10:

1/(x - √x)

1/(x - √x) = 1/{√x (√x - 1)}

Let √x – 1 = t

=> (1/2√x)dx = dt

=> (1/√x)dx = 2dt

Now, ʃ1/{√x (√x - 1)}dx = 2ʃ dt/t

= 2log|t| +C

= 2log|√x - 1| + C

Question 11:

x/√(x + 4), x > 0

Let x + 4 = t

=> dx = dt

Now, ʃ{ x/√(x + 4)}dx = ʃ{(t - 4)/√t} dt

= ʃ(√t - 4/√t) dt

= ʃ √t dt - ʃ 4/√t dt

= ʃ t1/2 dt - 4ʃ 1/t1/2 dt

= t3/2/(3/2) – 4t1/2/(1/2) + C

= 2t3/2/3 – 8t1/2 + C

= 2t * t1/2/3 – 8t1/2 + C

= 2t1/2(t - 12)/3 + C

= 2{(x + 4)1/2(x + 4 - 12)}/3 + C

= 2{√(x + 4)(x + 4 - 12)}/3 + C

Question 12:

(x3 - 1)1/3 x5

Let x3 – 1 = t

=> 3x2 dx = dt

=> x2 dx = dt/3

Now, ʃ(x3 - 1)1/3 x5 dx = ʃ(x3 – 1)1/3 x3 * x2 dx

= ʃ{t1/3 * (t + 1)}/3 dt

= (1/3)ʃ(t4/3 + t1/3) dt

= (1/3)[ʃ t4/3 dt + ʃ t1/3 dt]

= (1/3)[t7/3/(7/3) + t4/3/(4/3)] + C

= (1/3)[3t7/3/7 + 3t4/3/4] + C

= (x3 - 1)7/3/7 + (x3 - 1)4/3/4] + C

Question 13:

x2/(2 + 3x3)3

Let 2 + 3x3 = t

=> 9x2 dx = dt

=> x2 dx = dt/9

Now, ʃ x2/(2 + 3x3)3 = (1/9)ʃ dt/t3

= (1/9)ʃ t-3 dt

= (1/9) * t-2/(-2) + C

= (-1/18) * (1/t2) + C

= (-1/18) * [1/(2 + 3x3)2] + C

Question 14:

1/{x(log x)m}, x > 0

Let log x = t

=> dx/x = dt

Now, ʃ1/{x(log x)m} dx =ʃ dt/tm

= ʃ t-m dt

= t-m+1/(-m + 1) + C

= (log x)1-m/(1 - m) + C

Question 15:

x/(9 – 4x2)

Let 9 – 4x2 = t

=> -8x dx = dt

=> x dx = -dt/8

Now, ʃ [x/(9 – 4x2)]dx = (-1/8)ʃ dt/t

= (-1/8) log|t| + C

= (-1/8) log|9 – 4x2| + C

Question 16:

e2x + 3

Let 2x + 3 = t

=> 2 dx = dt

=> dx = dt/2

Now, ʃ e2x + 3 dx = ʃ et/2 dt

= et/2 + C

= e2x + 3/2 + C

Question 17:

x/ex2

Let x2 = t

=> 2x dx = dt

=> x dx = dt/2

Now, ʃx/ex = (1/2) ʃ(1/et) dt

= (1/2) ʃ e-t dt

= (1/2) * e-t/(-1) + C

= (-1/2) * e-x2 + C

= -1/2e-x2 + C

Question 18:

etan-1 x/(1 + x2)

Let tan-1 x = t

=> [1/(1 +x2)] dx = dt

Now, ʃ etan-1 x/(1 + x2) = ʃ et dt

= et + C

= etan-1 x + C

Question 19:

(e2x - 1)/(e2x + 1)

Given, (e2x - 1)/(e2x + 1)

Dividing numerator and denominator by ex, we get

[(e2x - 1)ex]/[(e2x + 1)/ex] = (ex - e-x)/(ex + e-x)

Let (ex + e-x) = t

=> (ex - e-x)dx = dt

Now, ʃ[(e2x - 1)/(e2x + 1)] dx = ʃ [(ex - e-x)/(ex + e-x)] dx

= ʃ dt/t

= log|t| + C

= log|ex + e-x| + C

Question 20:

(e2x – e-2x)/(e2x + e-2x)

Let e2x + e-2x = t

=> (2e2x - 2e-2x)dx = dt

=> (e2x - e-2x)dx = dt/2

Now, ʃ [(e2x – e-2x)/(e2x + e-2x)]dx = ʃ dt/2t

= (1/2)ʃ dt/t

= (1/2)log|t| + C

= (1/2)log|e2x + e-2x| + C

Question 21:

tan2(2x - 3)

tan2(2x - 3) = sec2(2x - 3) – 1

Let 2x – 3 = t

=> 2 dx = dt

=> dx = dt/2

Now, ʃ tan2(2x - 3) = ʃ [sec2(2x - 3) – 1] dx

= ʃ sec2(2x - 3) dx - ʃ dx

= (1/2) ʃ sec2 t dt - ʃ dx

= (1/2) tan t – x + C

= (1/2) tan (2x - 3) – x + C

Question 22:

sec2(7 – 4x)

Let 7 – 4x = t

=> -4dx = dt

=> dx = -dt/4

Now, ʃ sec2(7 – 4x) dx = (-1/4) ʃ sec2 t dt

= (-1/4) (tan t) + C

= (-1/4) tan (7 – 4x) + C

Question 23:

sin-1 x /√(1 – x2)

Let sin-1 x = t

=> [1/√(1 – x2)]dx = dt

Now, ʃ[sin-1 x /√(1 – x2)]dx = ʃ t dt

= t2/2 + C

= (sin-1 x)2/2 + C

Question 24:

(2 cos x – 3 sin x)/(6 cos x + 4 sin x)

(2 cos x – 3 sin x)/(6 cos x + 4 sin x) = (2 cos x – 3 sin x)/{2(3 cos x + 2 sin x)}

Let 3 cos x + 2 sin x = t

=> (-3 sin x + 2 cos x)dx = dt

=> (2 cos x - 3 sin x)dx = dt

Now, ʃ [(2 cos x – 3 sin x)/(6 cos x + 4 sin x)]dx = ʃ dt/2t

= (1/2) * ʃ dt/t

= (1/2) * log|t| + C

= (1/2) * log|2 sin x + 3 cos x| + C

Question 25:

1/[cos2 x(1 – tan x)2]

1/[cos2 x(1 – tan x)2] = sec2 x/(1 – tan x)2

Let 1 – tan x = t

=> -sec2 x dx = dt

=> sec2 x dx = -dt

Now, ʃ[sec2 x/(1 – tan x)2]dx = -ʃ dt/t2

= -ʃ t-2 dt

= - t-1/(-1) + C

= 1/t + C

= 1/(1 – tan x) + C

Question 26:

cos √x /√x

Let √x = t

=> (1/2√x)dx = dt

=> (1/√x)dx = 2 dt

Now, ʃ (cos √x /√x)dx = 2ʃ cos t dt

= 2 sin t + C

= 2 sin √x + C

Question 27:

√(sin 2x) * cos 2x

Let sin 2x = t

=> 2 cos 2x dx = dt

=> cos 2x dx = dt/2

Now, ʃ √(sin 2x) * cos 2x dx = (1/2)ʃ √t dt

= (1/2)ʃ t1/2 dt

= (1/2) * t3/2/(3/2)

= (1/3) * t3/2 + C

= (1/3) * (sin 2x)3/2 + C

Question 28:

cos x /√(1 + sin x)

Let 1 + sin x = t

=> cos x dx = dt

Now, ʃ[cos x /√(1 + sin x)]dx = ʃ dt/√t

= ʃ t-1/2 dt

= t1/2/(1/2) + C

= 2√t + C

= 2√(1 + sin x) + C

Question 29:

cot x * log sin x

Let log sin x = t

=> (1/sin x) * cos x dx = dt

=> cot x dx = dt

Now, ʃ cot x * log sin x dx = ʃ t dt

= t2/2 + C

= (log sin x)2/2 + C

Question 30:

sin x /(1 + cos x)

Let 1 + cos x = t

=> -sin x dx = dt

=> sin x dx = -dt

Now, ʃ[sin x /(1 + cos x)]dx = -ʃ dt/t

= -log|t| + C

= -log|1 + cos x| + C

Question 31:

sin x /(1 + cos x)2

Let 1 + cos x = t

=> -sin x dx = dt

=> sin x dx = -dt

Now, ʃ sin x /(1 + cos x)2 = -ʃ dt/t2

= -ʃ t-2 dt

= -t-1/(-1) + C

= 1/t + C

= 1/(1 + cos x) + C

Question 32:

1/(1 + cot x)

Let I = ʃ[1/(1 + cot x)]dx

= ʃ [1/(1 + cos x /sin x)]dx

= ʃ [sin x /(sin x + cos x)] dx

= (1/2) * ʃ [2 sin x /(sin x + cos x)] dx

= (1/2) * ʃ [(sin x + cos x + sin x – cos x)/(sin x + cos x)] dx

= (1/2) * ʃ [1 + (sin x – cos x)/(sin x + cos x)] dx

= (1/2) * ʃ dx + (1/2) * ʃ [(sin x – cos x)/(sin x + cos x)] dx

= x/2 + (1/2) * ʃ [(sin x – cos x)/(sin x + cos x)] dx

Let sin x + cos x = t

=> (cos x – sin x)dx = dt

=> -(sin x – cos x)dx = dt

=> (sin x – cos x)dx = -dt

Now, I = x/2 + (1/2) * ʃ [-dt/t] dx

=> I = x/2 + (1/2) * (-log |t|) + C

=> I = x/2 - (1/2) * log |sin x + cos x| + C

Question 33:

1/(1 – tan x)

Let I = ʃ [1/(1 – tan x)] dx

=> I = ʃ [1/(1 – sin x /cos x)] dx

=> I = ʃ [cos x /(cos x – sin x)] dx

=> I = (1/2)ʃ [2 cos x /(cos x – sin x)] dx

=> I = (1/2)ʃ [(cos x – sin x + cos x + sin x)/(cos x – sin x)] dx

=> I = (1/2)ʃ [1 + (cos x + sin x)/(cos x – sin x)] dx

=> I = (1/2)ʃ dx + (1/2)ʃ [(cos x + sin x)/(cos x – sin x)] dx

=> I = x/2 + (1/2)ʃ [(cos x + sin x)/(cos x – sin x)] dx

Let cos x – sin x = t

=> (-sin x – cos x)dx = dt

=> -(sin x + cos x)dx = dt

=> (sin x + cos x)dx = -dt

Now, I = x/2 + (1/2)ʃ (-dt/t)

=> I = x/2 - (1/2)ʃ dt/t

=> I = x/2 - (1/2) log |t| + C

=> I = x/2 - (1/2) log |cos x – sin x| + C

Question 34:

√(tan x)/(sin x * cos x)

Let I = ʃ [√(tan x)/(sin x * cos x)] dx

= ʃ [√(tan x * cos x)/(sin x * cos x * cos x)]dx

= ʃ [√(tan x)/(tan x * cos2 x)]dx

= ʃ [sec2 x /√(tan x)]dx

Let tan x = t

=> sec2 x dx = dt

Now, I = ʃ dt/√t

=> I = ʃ t-1/2 dt

=> I = 2√t + C

=> I = 2√( tan x) + C

Question 35:

(1 + log x)2/x

Let 1 + log x = t

=> dx/x = dt

Now, ʃ[(1 + log x)2/x]dx = ʃ t2 dt

= t3/3 + C

= (1 + log x)3/3 + C

Question 36:

{(x + 1)(x + log x)2}/x

{(x + 1)(x + log x)2}/x = {(x + 1)/x}(x + log x)2

= (1 + 1/x)(x + log x)2

Let x + log x = t

=> (1 + 1/x)dx = dt

Now, ʃ [(1 + 1/x)(x + log x)2]dx = ʃ t2 dt

= t3/3 + C

= (x + log x)3/3 + C

Question 37:

x3 * sin(tan-1 x4)/(1 + x8)

Let x4 = t

=> 4x3 dx = dt

=> x3 dx = dt/4

Now, ʃ[x3 * sin(tan-1 x4)/(1 + x8)] = (1/4)ʃ[sin(tan-1 t)/(1 + t2)] dt      ………………1

Let tan-1 t = u

=> dt/(1 + t2) = du

From equation 1, we get

=> ʃ[x3 * sin(tan-1 x4)/(1 + x8)] = (1/4)ʃ sin u du

= (1/4)(-cos u) + C

= (-1/4) * cos (tan-1 t) + C

= (-1/4) * cos (tan-1 x4) + C

Question 38:

ʃ [(10 x9 + 10x loge 10)/(x10 + 10x)]dx equals

(A) 10x – x10 + C                (B) 10x + x10 + C            (C) (10x – x10)-1 + C             (D) log (10x + x10) + C

Let x10 + 10x = t

=> (10 x9 + 10x loge 10)dx = dt

Now, ʃ [(10 x9 + 10x loge 10)/(x10 + 10x)]dx = ʃ dt/t

= log t + C

= log (x10 + 10x) + C

Hence, the correct answer is option D.

Question 39:

ʃ dx/(sin2 x * cos2 x)

1. tan x + cot x + C B. tan x – cot x + C
2. tan x * cot x + C D. tan x – cot 2x + C

Let I = ʃ dx/(sin2 x * cos2 x)

= ʃ [1/(sin2 x * cos2 x)]dx

= ʃ [(sin2 x + cos2 x)/(sin2 x * cos2 x)]dx

= ʃ [sin2 x/(sin2 x * cos2 x)]dx + ʃ [cos2 x/(sin2 x * cos2 x)]dx

= ʃ sec2 x dx + ʃ cosec2 x dx

= tan x – cot x + C

Hence, the correct answer is option B.

Exercise 7.3

Find the integrals of the functions in Exercises 1 to 22:

Question 1:

sin2(2x + 5)

sin2(2x + 5) = [1 – cos2(2x + 5)]/2 = [1 – cos(4x + 10)]/2

Now, ʃ sin2(2x + 5) dx = ʃ [1 – cos(4x + 10)]/2 dx

= (1/2) ʃ dx – (1/2) ʃ cos(4x + 10) dx

= x/2 – (1/2) [sin(4x + 10) /4] + C

= x/2 – sin(4x + 10) /8 + C

Question 2:

sin 3x * cos 4x

We know that

sin A * cos B = (1/2)[sin(A + B) + sin(A - B)]

Now, ʃ sin 3x * cos 4x dx = (1/2)ʃ [sin(3x + 4x) + sin(3x – 4x)] dx

= (1/2)ʃ [sin 7x + sin(-x)] dx

= (1/2)ʃ [sin 7x – sin x] dx

= (1/2)ʃ [sin 7x – sin x] dx

= (1/2)ʃ sin 7x dx – (1/2) ʃ sin x dx

= (1/2)(-cos 7x)/7 – (-cos x)/2 + C

= (-cos 7x)/14 + (cos x)/2 + C

Question 3:

cos 2x * cos 4x * cos 6x

We know that cos A * cos B = (1/2)[cos(A + B) + cos(A - B)]

Now, ʃ cos 2x * (cos 4x * cos 6x) = ʃ [cos 2x * (1/2){cos(4x + 6x) + cos(4x – 6x)}]dx

= ʃ [cos 2x * (1/2){cos 10 + cos(-2x)}]dx

= (1/2) ʃ [cos 2x * cos 10 + cos2 2x]dx

= (1/2) ʃ [(1/2){cos (2x + 10x) + cos (2x – 10) + (1 + cos 4x)/2]dx

= (1/4) ʃ [cos 12x + cos 8x + 1 + cos 4x]dx

= (1/4) [(sin 12x)/12 + (sin 8x)/8 + x + (sin 4x)/4] + C

Question 4:

sin3(2x + 1)

Let I = sin3(2x + 1)

Now, ʃ sin3(2x + 1) dx = ʃ [sin2(2x + 1) * sin(2x + 1)]dx

= ʃ [{1 - cos2(2x + 1)} * sin(2x + 1)]dx

Let cos(2x + 1) = t

=> -2sin(2x + 1) dx = dt

=> sin(2x + 1) dx = -dt/2

Now, I = (-1/2)ʃ [1 - t2]dt

= (-1/2)ʃ [t – t3/3] + C

= (-1/2)ʃ [cos(2x + 1) – { cos(2x + 1)}3/3] + C

= (-1/2)ʃ [cos(2x + 1) – { cos3(2x + 1)}/3] + C

Question 5:

sin3 x * cos3 x

Let I = ʃ sin3 x * cos3 x dx

= ʃ cos3 x * sin2 x * sin x dx

=> I = ʃ cos3 x * (1 - cos2 x) * sin x dx

Let cos x = t

=> -sin x dx = dt

=> sin x dx = -dt

Now, I = -ʃ t3 * (1 - t2) dt

= -ʃ (t3 – t5) dt

= -(t4/4– t6/6) + C

= -(cos4 x /4– cos6 x /6) + C

= -cos4 x /4+ cos6 x /6 + C

Question 6:

sin x * sin 2x * sin 3x

We know that sin A * sin B = (1/2)[cos(A - B) – cos(A + B)]

Now, ʃ sin x * sin 2x * sin 3x dx = ʃ [sin x * (1/2){cos(2x – 3x) – cos(2x + 3x)}]dx

= ʃ [sin x * (1/2){cos(-x) – cos 5x}]dx

= (1/2)ʃ [sin x * cos x – sin x * cos 5x]dx

= (1/2)ʃ (sin 2x /2)dx – (1/2)ʃ [sin x * cos 5x]dx

= (1/4)[-cos 2x /2] – (1/2)ʃ [(1/2){sin (x + 5x) + sin (x – 5x)}]dx

= -cos 2x /8 – (1/4)ʃ [sin 6x + sin (-4x)] dx

= -cos 2x /8 – (1/4)ʃ [sin 6x - sin 4x] dx

= -cos 2x /8 – (1/4) [-cos 6x /6 + cos 4x /4] + C

= -cos 2x /8 – (1/8) [-cos 6x /3 + cos 4x /2] + C

= (1/8)[cos 6x /3 –cos 4x /2 - cos 2x ] + C

Question 7:

sin 4x * sin 8x

We know that sin A * sin B = (1/2)[cos(A - B) – cos(A + B)]

So, ʃ sin 4x * sin 8x dx = (1/2)ʃ [cos(4x – 8x) – cos(4x + 8x)]dx

= (1/2)ʃ [cos(-4x) – cos 12x]dx

= (1/2)[sin 4x / 4 – sin 12x /12] + C

Question 8:

(1 – cos x)/(1 + cos x)

(1 – cos x)/(1 + cos x) = (2 sin2 x/2)/( 2 cos2 x/2) = tan2 x/2 = (sec2 x/2 - 1)

Now, ʃ[(1 – cos x)/(1 + cos x)]dx = ʃ(sec2 x/2 - 1) dx

= [(tan x/2)/(1/2) - x] + C

= 2(tan x/2) - x + C

Question 9:

cos x /(1 + cos x)

cos x /(1 + cos x) = (cos2 x/2 – sin2 x/2)/(2cos2 x/2)

= (1/2) * (1 – tan2 x/2)

So, ʃ [cos x /(1 + cos x)]dx = (1/2) ʃ (1 – tan2 x/2)dx

= (1/2) ʃ (1 – sec2 x/2 + 1)dx

= (1/2) ʃ (2 – sec2 x/2)dx

= (1/2)[2x – (tan x/2)/(1/2)] + C

= x – tan x/2 + C

Question 10:

sin4 x

sin4 x = sin2 x * sin2 x

= [(1 – cos 2x)/2] * [(1 – cos 2x)/2]

= (1 – cos 2x)2/4

= (1/4)[1 + cos2 2x – 2 cos 2x]

= (1/4)[1 + {(1 + cos 4x)/2} – 2 cos 2x]

= (1/4)[1 + 1/2 + (1/2)cos 4x – 2 cos 2x]

= (1/4)[3/2 + (1/2)cos 4x – 2 cos 2x]

Now, ʃ sin4 x dx = (1/4)ʃ [3/2 + (1/2)cos 4x – 2 cos 2x]dx

= (1/4)[3x/2 + (1/2)(sin 4x /4) – 2 (sin 2x /2)] + C

= (1/8)[3x + (sin 4x)/4 – 2 sin 2x] + C

= 3x/8 - (sin 2x)/4 + (sin 4x)/32 + C

Question 11:

cos4 2x

cos4 2x = (cos2 2x)2

= (1 + cos 4x)/2]2

= (1/4)[1 + cos2 4x + 2 cos 4x]

= (1/4)[1 + (1 + cos 8x)/2 + 2 cos 4x]

= (1/4)[1 + 1/2 + cos 8x /2 + 2 cos 4x]

= (1/4)[3/2 + cos 8x /2 + 2 cos 4x]

Now, ʃ cos4 2x dx = (1/4)ʃ [3/2 + cos 8x /2 + 2 cos 4x]dx

= (1/4)ʃ [3x/2 + sin 8x /16 + cos 4x /2] + C

= 3x/8 + sin 8x /64 + sin 4x /8 + C

Question 12:

sin2 x /(1 + cos x)

sin2 x /(1 + cos x) = (2 * sin x/2 * cos x/2)2/(2cos2 x/2)

= (4 * sin2 x/2 * cos2 x/2)/(2cos2 x/2)

= 2 * sin2 x/2

= 1 – cos x

Now, ʃ[sin2 x /(1 + cos x)]dx = ʃ (1 – cos x)dx

= x – sin x + C

Question 13:

(cos 2x – cos 2α)/(cos x – cos α)

(cos 2x – cos 2α)/(cos x – cos α)

= [-2 * sin(2x + 2α)/2 * sin(2x - 2α)/2]/[-2 * sin(x + α)/2 * sin(x - α)/2]

= [sin(x + α) * sin(x - α)]/[sin(x + α)/2 * sin(x - α)/2]

= [{2 * sin(x + α)/2 * cos(x + α)/2}* {2 * sin(x - α)/2 * cos(x - α)/2}]/[sin(x + α)/2 * sin(x - α)/2]

= 4 * cos(x + α)/2 * cos(x - α)/2

= 2 * [cos{(x + α)/2 + (x - α)/2} + cos{(x + α)/2 - (x - α)/2}]

= 2[cos x + cos α]

= 2 cos x + 2 cos α

Now, ʃ[(cos 2x – cos 2α)/(cos x – cos α)]dx = ʃ[2 cos x + 2 cos α]dx

= 2[sin x + x * cos α] + C

Question 14:

(cos x – sin x)/(1 + sin 2x)

(cos x – sin x)/(1 + sin 2x) = (cos x – sin x)/(sin2 x + cos2 x + 2 * sin x * cos x)

= (cos x – sin x)/(sin x + cos x)2

Let sin x + cos x = t

=> (cos x – sin x)dx = dt

Now, ʃ[(cos x – sin x)/(1 + sin 2x)]dx = ʃ[(cos x – sin x)/(sin x + cos x)2]dx

= ʃ dt/t2

= ʃ t-2 dt

= - t-1 + C

= -1/t + C

= -1/(sin x + cos x) + C

Question 15:

tan3 2x * sec 2x

tan3 2x * sec 2x = tan2 2x * tan 2x * sec 2x

= (sec2 2x – 1) * tan 2x * sec 2x

= sec2 2x * tan 2x * sec 2x – tan 2x * sec 2x

ʃ tan3 2x * sec 2x dx = ʃ sec2 2x * tan 2x * sec 2x dx – ʃ tan 2x * sec 2x dx

= ʃ sec2 2x * tan 2x * sec 2x dx – (sec 2x)/2 + C

Let sec 2x = t

=> 2 * sec 2x * tan 2x dx = dt

=> sec 2x * tan 2x dx = dt/2

Now, ʃ tan3 2x * sec 2x dx = ʃ sec2 2x * tan 2x * sec 2x dx – ʃ tan 2x * sec 2x dx

= (1/2)ʃ t2 dt – (sec 2x)/2 + C

= (1/2) * t3/3 – (sec 2x)/2 + C

= (sec 2x)3/6 – (sec 2x)/2 + C

Question 16:

tan4 x

tan4 x = tan2 x * tan2 x

= (sec2 x – 1) * tan2 x

= sec2 x * tan2 x – tan2 x

= sec2 x * tan2 x – (sec2 x – 1)

= sec2 x * tan2 x – sec2 x + 1

Now, ʃ tan4 x dx = ʃ [sec2 x * tan2 x – sec2 x + 1]dx

= ʃ sec2 x * tan2 x dx – ʃ sec2 x dx + ʃ dx

= ʃ sec2 x * tan2 x dx – tan x + x + C    ………..1

Let tan x = t

=> sec2 x dx = dt

From equation 1, we get

ʃ tan4 x dt = ʃ t2 dt – tan x + x + C

ʃ tan4 x dt = t3/3 – tan x + x + C

ʃ tan4 x dt = tan3 x /3 – tan x + x + C

Question 17:

(sin3 x + cos3 x)/(sin2 x * cos2 x)

(sin3 x + cos3 x)/(sin2 x * cos2 x) = sin3 x /(sin2 x * cos2 x) + cos3 x /(sin2 x * cos2 x)

= sin x /cos2 x + cos x /sin2 x

= tan x * sec x + cot x * cosec x

So, ʃ [(sin3 x + cos3 x)/(sin2 x * cos2 x)]dx = ʃ (tan x * sec x + cot x * cosec x) dx

= sec x – cosec x + C

Question 18:

(cos 2x + 2 sin2 x)/cos2 x

(cos 2x + 2 sin2 x)/cos2 x = (cos 2x + 1 – cos 2x)/cos2 x

= 1/cos2 x

= sec2 x

So, ʃ[(cos 2x + 2 sin2 x)/cos2 x]dx = ʃ sec2 x dx

= tan x + C

Question 19:

1/(sin x * cos3 x)

1/(sin x * cos3 x) = (sin2 x + cos2 x)/(sin x * cos3 x)

= sin2 x /(sin x * cos3 x) + cos2 x /(sin x * cos3 x)

= sin x /cos3 x + 1/(sin x * cos x)

= tan x * sec2 x + (1/ cos2 x)/{(sin x * cos x)/ cos2 x}

= tan x * sec2 x + sec2 x /tan x

Now, ʃ [1/(sin x * cos3 x)]dx = ʃ tan x * sec2 x dx + ʃ [sec2 x /tan x]dx

Let tan x = t

=> sec2 x dx = dt

So, ʃ [1/(sin x * cos3 x)]dx = ʃ t dt + ʃ dt/t

= t2/2 + log|t| + C

= (tan2 x)/2 + log|tan x| + C

Question 20:

cos 2x /(cos x + sin x)2

cos 2x /(cos x + sin x)2 = cos 2x /(cos2 x + sin2 x + 2 * sin x * cos x)

= cos 2x /(1 + sin 2x)

Now, ʃ [cos 2x /(cos x + sin x)2]dx = ʃ [cos 2x /(1 + sin 2x)]dx

Let 1 + sin 2x = t

=> 2 * cos 2x dx = dt

=> cos 2x dx = dt/2

So, ʃ [cos 2x /(cos x + sin x)2]dx = (1/2)ʃ dt/t

= (1/2) * log|t| + C

= (1/2) * log|1 + sin 2x| + C

= (1/2) * log|(sin x + cos x)2| + C

= (2/2) * log|(sin x + cos x)| + C

= log|(sin x + cos x)| + C

Question 21:

sin-1(cos x)

Let cos x = t

=> sin x = √(1 – t2)

And –sin x dx = dt

=> sin x dx = -dt

Now, ʃ sin-1(cos x) dx = ʃ sin-1 t * [-dt/√(1 – t2)]

= -ʃ [sin-1 t /√(1 – t2)]dt

Let sin-1 t = u

=> dt/√(1 – t2) = du

So, ʃ sin-1(cos x) dx = -ʃ u du

= -u2/2 + C

= -(sin-1 t)2/2 + C

= -[sin-1 (cos x)]2/2 + C    ……………1

We know that sin-1 x + cos-1 t = π/2

So, sin-1(cos x) = π/2 - cos-1(cos x) = π/2 – x

Put this value in equation 1, we get

So, ʃ sin-1(cos x) dx = -[ π/2 – x]2/2 + C

= (-1/2)[ π2/4 + x2 - πx]/2 + C

= -π2/8 - x2/2 + πx/2 + C

= πx/2 - x2/2 + (C - π2/8)

= πx/2 - x2/2 + C1

Question 22:

1/[cos(x - a) * cos(x - b)]

1/[cos(x - a) * cos(x - b)] = [1/sin(a - b)] * [sin(a - b)/{cos(x - a) * cos(x - b)}]

= [1/sin(a - b)] * [sin{(x - b) – (x - a)}/{cos(x - a) * cos(x - b)}]

= [1/sin(a - b)] * [{sin(x - b) *  cos(x - a) - cos(x - b) * sin(x - a)}

{cos(x - a) * cos(x - b)}]

= [1/sin(a - b)] * [tan(x - b) - tan(x - a)]

So, ʃ dx/[cos(x - a) * cos(x - b)] = [1/sin(a - b)] * ʃ [tan(x - b) - tan(x - a)]dx

= [1/sin(a - b)] * [-log|cos(x - b)| + log|cos(x - a)|] + C

= [1/sin(a - b)] * log|{cos(x - a)}/{ cos(x - b)}|  + C

Choose the correct answer in Exercises 23 and 24.

Question 23:

ʃ [(sin2 x – cos2 x)/(sin2 x * cos2 x)]dx is equal to

1. tan x + cot x + C B. tan x + cosec x + C
2. -tan x + cot x + C D. tan x + sec x + C

Given, ʃ [(sin2 x – cos2 x)/(sin2 x * cos2 x)]dx

= ʃ [(sin2 x)/(sin2 x * cos2 x)]dx - ʃ [cos2 x)/(sin2 x * cos2 x)]dx

= ʃ sec2 x dx - ʃ cosec2 x dx

= tan x + cot x + C

Hence, the correct answer is option A.

Question 24:

ʃ[{ex(1 + x)}/cos2(ex x)] dx equals

1. –cot(e xx)+ C B. tan(xex) + C C. tan(ex) + C                  D. cot(ex) + C

Let ex x = t

=> (ex x + ex * 1)dx = dt

=> ex (x + 1)dx = dt

So, ʃ[{ex(1 + x)}/cos2(ex x)] dx = ʃ [dt/cos2 t]

= ʃ sec2 t dt

= tan t + C

= tan (ex x) + C

Hence, the correct answer is option B.

Exercise 7.4

Integrate the functions in Exercises 1 to 23.

Question 1:

3x2/(x6 + 1)

Let x3 = t

=> 3x2 dx = dt

So, ʃ [3x2/(x6 + 1)]dx = ʃdt/(t2 + 1)

= tan-1 t + C

= tan-1 (x3) + C

Question 2:

1/√(1 + 4x2)

1/√(1 + 4x2) = 1/√{1 + (2x)2}

Let 2x = t

=> 2 dx = dt

So, ʃ dx/√(1 + 4x2) = (1/2) * ʃ dt/√(1 + t2)

= (1/2) * log|t + √(t2 + 1)| + C

= (1/2) * log|2x + √(4x2 + 1)| + C

Question 3:

1/√{(2 - x)2+ 1}

Let 2 – x = t

=> -dx = dt

=> dx = -dt

So, ʃ dx/√{(2 - x)2+ 1} = -ʃ dx/√(t2+ 1)

= -log|t + √(t2+ 1)| + C

= -log|(2 - x) + √{(2 - x)2+ 1}| + C

= log|1/[(2 - x) + √{(2 - x)2+ 1}]| + C

= log|1/[(2 - x) + √(x2 – 4x + 5]| + C

Question 4:

1/√(9 - 25x2)

1/√(9 - 25x2) = 1/√{9 – (5x)2}

Let 5x = t

=> 5 dx = dt

=> dx = dt/5

So, ʃ dx/√(9 - 25x2) = (1/5) * ʃ dx/√(9 - t2)

= (1/5) * ʃ dx/√(32 - t2)

= (1/5) * sin-1 (t/3) + C

= (1/5) * sin-1 (5x/3) + C

Question 5:

3x/(1 + 2x4)

3x/(1 + 2x4) = 3x/(1 + (√2x2)2}

Let √2x2 = t

=> 2√2x dx = dt

=> x dx = dt/2√2

So, ʃ [3x/(1 + 2x4)]dx = (3/2√2) * ʃ dt/(1 + t2)

= (3/2√2) * tan-1 t + C

= (3/2√2) * tan-1 (√2x2) + C

Question 6:

x2/(1 + x6)

x2/(1 + x6) = x2/(1 + (x3)2}

Let x3 = t

=> 3x2 dx = dt

=> x2 dx = dt/3

So, ʃ [x2/(1 + x6)]dx = (1/3) * ʃ dt/(1 - t2)

= (1/3) * [(1/2) * log|(1 + t)/(1 - t)|] + C

= (1/6) * log|(1 + x3)/(1 – x3)|] + C

Question 7:

(x - 1)/√(x2 - 1)

ʃ[(x - 1)/√(x2 - 1)]dx = ʃ[x/√(x2 - 1)]dx - ʃ[dx/√(x2 - 1)]dx  …………1

Consider, ʃ[x/√(x2 - 1)]dx

Let x2 – 1 = t

=> 2x dx = dt

=> x dx = dt/2

So, ʃ[x/√(x2 - 1)]dx = (1/2) * ʃ dt/√t

= (1/2) * ʃ t-1/2 dt

= (1/2) * t1/2/(1/2)

= t1/2

= √t

= √(x2 - 1)

From equation 1, we get

ʃ[(x - 1)/√(x2 - 1)]dx = ʃ[x/√(x2 - 1)]dx - ʃ[dx/√(x2 - 1)]dx

=> ʃ[(x - 1)/√(x2 - 1)]dx = √(x2 - 1) - log[x + √(x2 - 1)] + C

Question 8:

x2/√(x6 + a6)

x2/√(x6 + a6) = x2/√{(x3)2 + (a3)2}

Let x3 = t

=> 3x2 dx = dt

=> x2 dx = dt/3

So, ʃ [x2/√(x6 + a6)]dx = ʃ dt/√{t2 + (a3)2}

= (1/3) * log[t + √(t2 + a6)] + C

= (1/3) * log[x3 + √(x6 + a6)] + C

Question 9:

sec2 x/√(tan2 x + 4)

Let tan x = t

=> sec2 x dx = dt

So, ʃ[sec2 x/√(tan2 x + 4)]dx = ʃ[dt/√(t2 + 22)]

= log[t + √(t2 + 4)] + C

= log[tan x + √(tan2 x + 4)] + C

Question 10:

1/√(x2 + 2x + 2)

1/√(x2 + 2x + 2) = 1/√{(x + 1)2 + 12}

Let x + 1 = t

=> dx = dt

So, ʃ dx/√(x2 + 2x + 2) = ʃ dt/√(t2 + 12)

= log[t + √(t2 + 1)] + C

= log[(x + 1) + √{(x + 1)2 + 1)}] + C

= log[(x + 1) + √(x2 + 2x + 2)] + C

Question 11:

1/√(9x2 + 6x + 5)

1/√(9x2 + 6x + 5) = 1/√{(3x + 1)2 + 22}

Let 3x + 1 = t

=> 3 dx = dt

=> dx = dt/3

So, ʃ dx/√(x2 + 2x + 2) = (1/3) * ʃ dt/√(t2 + 22)

= (1/3) * [(1/2) * tan-1(t/2)] + C

= (1/6) * tan-1{(3x + 1)/2} + C

Question 12:

1/√(7 – 6x – x2)

7 – 6x – x2 =-(x2 + 6x – 7)

= -(x2 + 6x + 9 – 9 – 7)

= -(x2 + 6x + 9 – 16)

= -{(x + 3)2 – 42}

= 42 - (x + 3)2

So, ʃ dx/√(7 – 6x – x2) = ʃ dx/√{42 - (x + 3)2}

Let x + 3 = t

=> dx = dt

Now, ʃ dx/√(7 – 6x – x2) = ʃ dt/√(42 - t2)

= sin-1 (t/4) + C

= sin-1 {(x + 3)/4} + C

Question 13:

1/√{(x - 1)(x - 2)}

(x - 1)(x - 2) = x2 – 3x + 2

= x2 – 3x + 9/4 – 9/4 + 2

= (x – 3/2)2 – 1/4

= (x – 3/2)2 – (1/2)2

So, ʃ dx/√{(x - 1)(x - 2)} = ʃ dx/√{(x – 3/2)2 – (1/2)2}

Let x – 3/2 = t

=> dx = dt

Now, ʃ dx/√{(x - 1)(x - 2)} = ʃ dx/√{t2 – (1/2)2}

= log|t + {t2 – (1/2)2}| + C

= log|(x – 3/2) + √{(x – 3/2)2 – (1/2)2}| + C

= log|(x – 3/2) + √(x2 – 3x + 2)| + C

Question 14:

1/√(8 + 3x – x2)

8 + 3x – x2 =-(x2 - 3x – 8)

= -(x2 - 3x + 9/4 – 9/4 – 8)

= -{(x – 3/2)2 – 41/4}

= -{(x – 3/2)2 – (√41/2)2}

= (√41/2)2 - (x – 3/2)2

So, ʃ dx/√(8 + 3x – x2) = ʃ dx/√{(√41/2)2 - (x – 3/2)2}

Let x – 3/2 = t

=> dx = dt

Now, ʃ dx/√(8 + 3x – x2) = ʃ dt/√{(√41/2)2 - t2}

= sin-1 {t/(√41/2)} + C

= sin-1 {(x – 3/2)/(√41/2)} + C

= sin-1 {(2x – 3)/√41} + C

Question 15:

1/√{(x - a)(x - b)}

(x - a)(x - a) = x2 – (a + b)x + ab

= x2 – (a + b)x + (a + b)2/4 – (a + b)2/4 + ab

= x2 – (a + b)x + (a + b)2/4 – {(a + b)2 + 4ab}/4

= {x – (a + b)/2}2 – (a - b)2/4

So, ʃ dx/√{(x - a)(x - b)} = ʃ dx/√[{x – (a + b)/2}2 – {(a - b)/2}2]

Let x – (a + b)/2 = t

=> dx = dt

Now, ʃ dx/√{(x - a)(x - b)} = ʃ dx/√[t2 – {(a - b)/2}2]

= log|t + [t2 – {(a - b)/2}2| + C

= log|{x – (a + b)/2} + √{(x - a)(x - b)}| + C

Question 16:

(4x + 1)/√(2x2 + x - 3)

Let 4x + 1 = A * d(2x2 + x - 3)/dx + B

=> 4x + 1 = A(4x + 1) + B

=> 4x + 1 = 4Ax + A + B

Equating the coefficients of x and constant term on both sides, we get

4A = 4

=> A = 1

A + B = 1

=> 1 + B = 1

=> B = 0

Let 2x2 + x – 3 = t

=> (4x + 1)dx = dt

Now, ʃ[(4x + 1)/√(2x2 + x - 3)]dx = ʃ dt/√t

= 2√t + C

= 2√(2x2 + x - 3) + C

Question 17:

(x + 2)/√(x2 - 1)

Let x + 2 = A * d(x2 - 1)/dx + B

=> x + 2 = A(2x) + B   ………………1

Equating the coefficients of x and constant term on both sides, we get

2A = 1

=> A = 1/2

B = 2

From equation 1, we get

x + 2 = (2x)/2 + 2

Now, ʃ[(x + 2)/√(x2 - 1)]dx = ʃ [{(2x)/2 + 2)/√(x2 - 1)]dx

= (1/2) * ʃ [2x)/√(x2 - 1)]dx + ʃ [2/√(x2 - 1)]dx      ………..2

In (1/2) * ʃ [2x)/√(x2 - 1)]dx,

Let x2 – 1 = t

=> 2x dx = dt

So, (1/2) * ʃ [2x)/√(x2 - 1)]dx = (1/2) * ʃ dt/√t

= (1/2) * 2√t

= √t

= √(x2 - 1)

Again, ʃ [2/√(x2 - 1)]dx = 2 ʃ dx/√(x2 - 1)

= 2 * log|x + √(x2 - 1)|

From equation 2, we get

Now, ʃ[(x + 2)/√(x2 - 1)]dx = √(x2 - 1)] + 2 * log|x + √(x2 - 1)| + C

Question 18:

(5x - 2)/(1 + 2x + 3x2)

Let (5x - 2) = A * d(1 + 2x + 3x2)/dx + B

=> 5x – 2 = A(2 + 6x) + B

Equating the coefficients of x and constant term on both sides, we get

6A = 5

=> A = 5/6

2A + B = -2

=> 2 * 5/6 + B = -2

=> 5/3 + B = -2

=> B = -2 – 5/3

=> B = -11/3

So, 5x – 2 = 5(2 + 6x)/6 – 11/3

Now, ʃ [(5x - 2)/(1 + 2x + 3x2)]dx

= ʃ [{5(2 + 6x)/6 – 11/3}/(1 + 2x + 3x2)]dx

= (5/6) * ʃ [(2 + 6x)/(1 + 2x + 3x2)]dx – (11/3) * ʃ dx/(1 + 2x + 3x2)

Let I1 = ʃ [(2 + 6x)/(1 + 2x + 3x2)]dx and I2 = ʃ dx/(1 + 2x + 3x2)

So, ʃ [(5x - 2)/(1 + 2x + 3x2)]dx = 5I1/6 – 11I2/3   ………………1

Now, I1 = ʃ [(2 + 6x)/(1 + 2x + 3x2)]dx

Let 1 + 2x + 3x2 = t

=> (2 + 6x)dx = dt

So, I1 = ʃ dt/t

=> I1 = log|t|

=> I1 = log|1 + 2x + 3x2|     …………….2

I2 = ʃ dx/(1 + 2x + 3x2

Given, 1 + 2x + 3x2 = 3(x2 + 2x/3 + 1/3)

= 3(x2 + 2x/3 + 1/9 – 1/9 + 1/3)

= 3{(x + 1/3)2 + 2/9}

= 3{(x + 1/3)2 + (√2/3)2}

I2 = ʃ dx/(1 + 2x + 3x2

= ʃ dx/[3{(x + 1/3)2 + (√2/3)2}]

= (1/3) * ʃ dx/[{(x + 1/3)2 + (√2/3)2}]

= (1/3) * {1/(√2/3)} * tan-1{(x + 1/3)/(√2/3)}

= (1/3) * (3/√2) * tan-1{(3x + 1)/√2}

= (1/√2) * tan-1{(3x + 1)/√2}

From equation 1, we get

ʃ [(5x - 2)/(1 + 2x + 3x2)]dx = (5/6) * log|1 + 2x + 3x2| - (11/3) * (1/√2) * tan-1{(3x + 1)/√2} + C

=> ʃ [(5x - 2)/(1 + 2x + 3x2)]dx = (5/6) * log|1 + 2x + 3x2| - (11/3√2) * tan-1{(3x + 1)/√2} + C

Question 19:

(6x + 7)/√{(x - 5)(x - 4)}

Given, (6x + 7)/√{(x - 5)(x - 4)} = (6x + 7)/√(x2 - 9x + 20)

Let 6x + 7 = A * d(x2 - 9x + 20)/dx + B

=> 6x + 7 = A * (2x - 9)/dx + B

Equating the coefficients of x and constant term, we obtain

2A = 6

=> A = 3

−9A + B = 7

=> B = 34

So, 6x + 7 = 3 (2x − 9) + 34

Now, ꭍ[(6x + 7)/√(x2 - 9x + 20)]dx = ꭍ [{3(2x - 9) + 34}/√(x2 - 9x + 20)]dx

= 3 * ꭍ [(2x - 9)/√(x2 - 9x + 20)]dx + ꭍ dx/√(x2 - 9x + 20)

Let I1 = ꭍ [(2x - 9)/√(x2 - 9x + 20)]dx and I2 = ꭍ dx/√(x2 - 9x + 20)

So, ꭍ[(6x + 7)/√(x2 - 9x + 20)]dx = 3I1 + 34I2     ……………1

Now, consider I1 = ꭍ [(2x - 9)/√(x2 - 9x + 20)]dx

Let x2 - 9x + 20 = t

=> (2x - 9)dx = dt

Now, I1 = ꭍ dt/√t

=> I1 = 2√t

=> I1 = 2√( x2 - 9x + 20)

Again consider I2 = ꭍ dx/√(x2 - 9x + 20)

=> I2 = ꭍ dx/√(x2 - 9x + 20 + 81/4 – 81/4)

=> I2 = ꭍ dx/√{(x2 - 9x + 81/4) + 20 – 81/4}

=> I2 = ꭍ dx/√{(x – 9/2)2 - 1/4}

=> I2 = ꭍ dx/√{(x – 9/2)2 – (1/2)2}

=> I2 = log|(x – 9/2) + √(x2 - 9x + 20)|

From equation 1, we get

ꭍ[(6x + 7)/√(x2 - 9x + 20)]dx = 3 * 2√( x2 - 9x + 20) + 34 * log|(x – 9/2) + √(x2 - 9x + 20)|+ C

=> ꭍ[(6x + 7)/√(x2 - 9x + 20)]dx = 6√( x2 - 9x + 20) + 34 * log|(x – 9/2) + √(x2 - 9x + 20)| + C

Question 20:

(x + 2)/√(4x – x2)

Let x + 2 = A * d(4x – x2)/dx + B

=> x + 2 = A * (4 – 2x) + B

Equating the coefficients of x and constant term on both sides, we get

-2A = 1

=> A = -1/2

And 4A + B = 2

=> B = 2

So, x + 2 = -(4 – 2x)/2 + 4

Now, ꭍ[(x + 2)/√(4x – x2)]dx = ꭍ[{-(4 – 2x)/2 + 4}/√(4x – x2)]dx

= (-1/2) * ꭍ[(4 – 2x)/√(4x – x2)]dx + 4 * ꭍ[dx/√(4x – x2)]

Let I1 = ꭍ[(4 – 2x)/√(4x – x2)]dx and I2 = ꭍ[dx/√(4x – x2)]

So, ꭍ[(x + 2)/√(4x – x2)]dx =-I1/2 + 4I2    ………………1

Consider I1 = ꭍ[(4 – 2x)/√(4x – x2)]dx

Let 4x – x2 = t

=> (4 – 2x)dx = dt

So, I1 = ꭍ dt/√t

=> I1 = 2√t

=> I1 = 2√(4x – x2)

Again, consider I2 = ꭍ[dx/√(4x – x2)]

=> I2 = ꭍ[dx/√{-(x2 – 4x)}]

=> I2 = ꭍ[dx/√{-(x2 – 4x + 4 - 4)}]

=> I2 = ꭍ[dx/√{-(x – 2)2 + 4}]

=> I2 = ꭍ[dx/√{-(x – 2)2 + 22}]

=> I2 = ꭍ[dx/√{22 - (x – 2)2}]

=> I2 = sin-1{(x – 2)/2}

From equation 1, we get

ꭍ [(x + 2)/√(4x – x2)]dx =-{2√(4x – x2)}/2 + 4 * sin-1{(x – 2)/2} + C

=> ꭍ [(x + 2)/√(4x – x2)]dx =-(4x – x2) + 4 * sin-1{(x – 2)/2} + C

Question 21:

(x + 2)/√(x2 + 2x + 3)

ꭍ[(x + 2)/√(x2 + 2x + 3)]dx = (1/2) * ꭍ [2(x + 2)/√(x2 + 2x + 3)]dx

= (1/2) * ꭍ [(2x + 4)/√(x2 + 2x + 3)]dx

= (1/2) * ꭍ [(2x + 2)/√(x2 + 2x + 3)]dx + (1/2) * ꭍ [2/√(x2 + 2x + 3)]dx

= (1/2) * ꭍ [(2x + 2)/√(x2 + 2x + 3)]dx + ꭍ dx/√(x2 + 2x + 3)

Let I1 = ꭍ [(2x + 2)/√(x2 + 2x + 3)]dx and I2 = ꭍ dx/√(x2 + 2x + 3)

So, ꭍ [(x + 2)/√(x2 + 2x + 3)]dx = I1/2 + I2    …………….1

Now, consider I1 = ꭍ [(2x + 2)/√(x2 + 2x + 3)]dx

Let x2 + 2x + 3 = t

=> (2x + 2)dx = dt

So, I1 = ꭍ dt/√t

=> I1 = 2√t

=> I1 = 2√(x2 + 2x + 3)

Again consider I2 = ꭍ dx/√(x2 + 2x + 3)

=> I2 = ꭍ dx/√(x2 + 2x + 1 + 2)

=> I2 = ꭍ dx/√{(x + 1)2 + 2}

=> I2 = ꭍ dx/√{(x + 1)2 + (√2)2}

=> I2 = log|(x + 1) + √(x2 + 2x + 3)|

From equation 1, we get

ꭍ [(x + 2)/√(x2 + 2x + 3)]dx =[2√(x2 + 2x + 3)]/2 + log|(x + 1) + √(x2 + 2x + 3)| + C

=> ꭍ [(x + 2)/√(x2 + 2x + 3)]dx =√(x2 + 2x + 3)] + log|(x + 1) + √(x2 + 2x + 3)| + C

Question 22:

(x + 3)/√(x2 – 2x - 5)

Let x + 3 = A * d(x2 – 2x - 5)/dx + B

=> x + 3 = A(2x - 2) + B

Equating the coefficients of x and constant term on both sides, we get

2A = 1

=> A = 1/2

And -2A + B = 3

=> B = 4

So, x + 3 = (2x - 2)/2 + 4

Now, ꭍ [(x + 3)/√(x2 – 2x - 5)]dx = ꭍ [{(2x - 2)/2 + 4}/√(x2 – 2x - 5)]dx

= (1/2) * ꭍ [(2x - 2) /√(x2 – 2x - 5)]dx + 4 * ꭍ dx/√(x2 – 2x - 5)

Let I1 = ꭍ [(2x - 2) /√(x2 – 2x - 5)]dx and I2 = ꭍ dx/√(x2 – 2x - 5)

So, ꭍ [(x + 3)/√(x2 – 2x - 5)]dx = I1/2 + 4I2   …………..1

Now, consider I1 = ꭍ [(2x - 2)/√(x2 – 2x - 5)]dx

Let x2 – 2x – 5 = t

=> (2x - 2)dx = dt

So, I1 = ꭍ dt/√t

=> I1 = 2√t

=> I1 = 2√(x2 – 2x – 5)

Again consider I2 = ꭍ dx/√(x2 – 2x - 5)

=> I2 = ꭍ dx/√{(x – 1)2 – 6}

=> I2 = ꭍ dx/√{(x – 1)2 – √(6)2}

=> I2 = (1/2√6) * log{(x – 1 - √6)/(x – 1 + √6)}

From equation 1, we get

ꭍ [(x + 3)/√(x2 – 2x - 5)]dx = [2√(x2 – 2x – 5)]/2 + 4 * (1/2√6) * log{(x – 1 - √6)/(x – 1 + √6)}

=> ꭍ [(x + 3)/√(x2 – 2x - 5)]dx = √(x2 – 2x – 5)] + (2/√6) * log{(x – 1 - √6)/(x – 1 + √6)}

Question 23:

(5x + 3)/√(x2 + 4x + 10)

Let (5x + 3) = A * d(x2 + 4x + 10)/dx + B

=> (5x + 3) = A(2x + 4) + B

Equating the coefficients of x and constant term on both sides, we get

2A = 5

=> A = 5/2

And 4A + B = 3

=> B = -7

So, (5x + 3) = 5(2x + 4)/2 – 7

Now, ꭍ [(5x + 3)/√(x2 + 4x + 10)]dx = ꭍ [{5(2x + 4)/2 – 7}/√(x2 + 4x + 10)]dx

= (5/2) * ꭍ [(2x + 4)/√(x2 + 4x + 10)]dx - 7 * ꭍ dx/√(x2 + 4x + 10)

Let I1 = ꭍ [(2x + 4)/√(x2 + 4x + 10)]dx and I2 = ꭍ dx/√(x2 + 4x + 10)

So, ꭍ [(5x + 3)/√(x2 + 4x + 10)]dx = 5I1/2 – 7I2    ……………..1

Now, consider I1 = ꭍ [(2x + 4)/√(x2 + 4x + 10)]dx

Let x2 + 4x + 10 = t

=> (2x + 4)dx = dt

So, I1 = ꭍ dt/√t

=> I1 = 2√t

=> I1 = 2√(x2 + 4x + 10)

Again consider I2 = ꭍ dx/√(x2 + 4x + 10)

=> I2 = ꭍ dx/√(x2 + 4x + 4 + 6)

=> I2 = ꭍ dx/√{(x + 2)2 + 6}

=> I2 = ꭍ dx/√{(x + 2)2 + (√6)2}

=> I2 = log{(x + 2) + (x2 + 4x + 10)}

From equation 1, we get

ꭍ [(5x + 3)/√(x2 + 4x + 10)]dx = 5[2√(x2 + 4x + 10)]/2 – 7 * log{(x + 2) + (x2 + 4x + 10)}

=> ꭍ [(5x + 3)/√(x2 + 4x + 10)]dx = 5 * √(x2 + 4x + 10) – 7 * log{(x + 2) + (x2 + 4x + 10)}

Question 24:

ʃ dx/(x2 + 2x + 2) equals

1. x * tan-1(x + 1) + C B. tan-1(x + 1) + C
2. (x + 1) * tan-1(x + 1) + C D. tan-1(x + 1) + C

ʃ dx/(x2 + 2x + 2) = ʃ dx/{(x2 + 2x + 1) + 1}

= ʃ dx/{(x + 1)2 + 12}

= tan-1(x + 1) + C

Hence, the correct answer is option B.

Question 25:

ʃ dx/√(9x - 4x2) equals

1. (1/9) * sin-1{(9x - 8)/8} + C B. (1/2) * sin-1{(8x - 9)/9} + C
2. (1/3) * sin-1{(9x - 8)/8} + C D. (1/2) * sin-1{(9x - 8)/9} + C

Given, ʃ dx/√(9x - 4x2) = ʃ dx/√{-4(x2 – 9x/4)}

= ʃ dx/√{-4(x2 – 9x/4 + 81/64 – 81/64)}

= ʃ dx/√[-4{(x – 9/8)2 –(9/8)2}]

= (1/2) * ʃ dx/√{(9/8)2 - (x – 9/8)2}

= (1/2) * sin-1{(x – 9/8)/(9/8)} + C

= (1/2) * sin-1{(8x – 9)/9} + C

Hence, the correct answer is option B.

Exercise 7.5

Integrate the rational functions in Exercises 1 to 21.

Question 1:

x/{(x + 1)(x + 2)}

Let x/{(x + 1)(x + 2)} = A/(x + 1) + B/(x + 2)

=> x = A(x + 2) + B(x + 1)

Equating the coefficients of x and constant term, we obtain

A + B = 1

2A + B = 0

On solving, we get

A = −1 and B = 2

So, x/{(x + 1)(x + 2)} = -1/(x + 1) + 2/(x + 2)

Now, ꭍ x dx/{(x + 1)(x + 2)} = ꭍ-dx/(x + 1) + ꭍ2dx/(x + 2)

=> ꭍ x dx/{(x + 1)(x + 2)} = (-1) * ꭍ dx/(x + 1) + 2 * ꭍ dx/(x + 2) + C

=> ꭍ x dx/{(x + 1)(x + 2)} = (-1) * log|x + 1| + 2 * log|x + 2| + C

=> ꭍ x dx/{(x + 1)(x + 2)} = log(x + 2)2 – log(x + 1) + C

=> ꭍ x dx/{(x + 1)(x + 2)} = log[(x + 2)2/(x + 1)] + C

Question 2:

1/(x2 - 9)

1/(x2 - 9) = 1/{(x + 3)(x - 3)}

Let 1/{(x + 3)(x - 3)} = A/(x + 3) + B/(x - 3)

=> 1 = A(x - 3) + B(x + 3)

Equating the coefficients of x and constant term, we obtain

A + B = 0

-3A + 3B = 1

On solving, we get

A = −1/6 and B = 1/6

So, 1/{(x + 3)(x - 3)} = -1/6(x + 3) + 1/6(x - 3)

Now, ꭍ dx/(x2 - 9) = ꭍ-dx/6(x + 3) + ꭍdx/6(x - 3)

=> ꭍ dx/(x2 - 9) = (-1/6) * ꭍ dx/(x + 3) + (1/6)ꭍ dx/(x - 3) + C

=> ꭍ dx/(x2 - 9) = (-1/6) * log| x + 3| + (1/6) * log| x - 3| + C

=> ꭍ dx/(x2 - 9) = (1/6) * log| x - 3| + (1/6) * log| x + 3| + C

=> ꭍ dx/(x2 - 9) = (1/6) * log|(x – 3)/(x + 3)| + C

Question 3:

(3x - 1)/{(x - 1)(x - 2)(x - 3)}

Let (3x - 1)/{(x - 1)(x - 2)(x - 3)} = A/(x - 1) + B/(x - 2) + C/(x - 3)

=> 3x - 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)

Equating the coefficients of x2, x and constant term, we get

A + B + C = 0

-5A – 4B – 3C = 3

6A + 3B + 2C = – 1

Solving these equations, we get

A = 1, B = −5, and C = 4

So, (3x - 1)/{(x - 1)(x - 2)(x - 3)} = 1/(x - 1) – 5/(x - 2) + 4/(x - 3)

Now, ꭍ [(3x - 1)/{(x - 1)(x - 2)(x - 3)}]dx = ꭍ dx/(x - 1) – ꭍ 5dx/(x - 2) + ꭍ 4dx/(x - 3)

=> ꭍ [(3x - 1)/{(x - 1)(x - 2)(x - 3)}]dx = ꭍ dx/(x - 1) – 5 ꭍ dx/(x - 2) + 4 ꭍ dx/(x - 3)

=> ꭍ [(3x - 1)/{(x - 1)(x - 2)(x - 3)}]dx = log|x – 1| – 5 log|x – 2| + 4 log|x – 3| + C

Question 4:

x/{(x - 1)(x - 2)(x - 3)}

Let x/{(x - 1)(x - 2)(x - 3)} = A/(x - 1) + B/(x - 2) + C/(x - 3)

=> x = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)

Equating the coefficients of x2, x and constant term, we get

A + B + C = 0

-5A – 4B – 3C = 1

6A + 3B + 2C = 0

Solving these equations, we get

A = 1/2, B = −2, and C = 3/2

So, x/{(x - 1)(x - 2)(x - 3)} = 1/2(x - 1) – 2/(x - 2) + 3/2(x - 3)

Now, ꭍ [x/{(x - 1)(x - 2)(x - 3)}]dx = ꭍ dx/2(x - 1) – ꭍ dx/2(x - 2) + ꭍ 3dx/2(x - 3)

=> ꭍ [x/{(x - 1)(x - 2)(x - 3)}]dx = (1/2) * ꭍ dx/(x - 1) – 2 ꭍ dx/(x - 2) + (3/2) * ꭍ dx/(x - 3)

=> ꭍ [(3x - 1)/{(x - 1)(x - 2)(x - 3)}]dx = (1/2) * log|x – 1| – 2 log|x – 2| + (3/2) * log|x – 3| + C

Question 5:

2x/(x2 + 3x + 2)

Let 2x/(x2 + 3x + 2) = A/(x +1) + B/(x + 2)

=> 2x = A(x + 2) + B(x + 1)

=> 2x = (A + B)x + (2A + B)

Equating the coefficients of x and constant term, we get

A + B = 2

2A + B = 0

Solving these equations, we get

A = -2 and B = 4

So, 2x/(x + 1)(x + 2) = -2/(x +1) + 4/(x + 2)

Now, ʃ [2x/(x + 1)(x + 2)] dx = ʃ [-2/(x +1)] dx + ʃ [4/(x + 2)] dx

= -2log|x + 1| + 4log|x + 2| + C

= 4log|x + 2| - 2log|x + 1| + C

Question 6:

(1 – x2)/{x(1 – 2x)}

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (1 − x2) by x(1 − 2x), we get

(1 – x2)/{x(1 – 2x)} = 1/2 + (1/2) [(2 – x)/{x(1 – 2x)}]   ……………..1

Let (2 – x)/{x(1 – 2x)} = A/x + B/(1 – 2x)

=> 2 – x = A(1 – 2x) + Bx

=> 2 – x = (-2A + B)x + A

Equating the coefficients of x and constant term, we obtain

-2A + B = -1

A = 2

Solving these equations, we get

A= 2 and B = 3

(2 – x)/{x(1 – 2x)} = 2/x + 3/(1 – 2x)

From equation 1, we get

(1 – x2)/{x(1 – 2x)} = 1/2 + (1/2) [2/x + 3/{x(1 – 2x)}

Now, ʃ [(1 – x2)/{x(1 – 2x)}]dx = ʃ [1/2 + (1/2){2/x + 3(1 – 2x)}] dx

= (1/2 )ʃ dx + ʃ dx/x + (3/2)ʃ dx/(1 – 2x)

= x/2 + log|x| - (3/4) log|1 – 2x| + C

Question 7:

x/{(x2 + 1)(x - 1)}

Let x/{(x2 + 1)(x - 1)} = (Ax + B)/(x2 + 1) + C/(x - 1)

=> x = (Ax + B)(x - 1) + C(x2 + 1)

=> x = Ax2 – Ax + Bx – B + Cx2 + C

Equating the coefficients of x2, x and constant terms, we get

A + C = 0

-A + B = 1

-B + C = 0

On solving these equations, we get

A = -1/2, B = 1/2 and C = 1/2

From equation 1, we get

x/{(x2 + 1)(x - 1)} = (-x/2 + 1/2)/(x2 + 1) + 1/2(x - 1)

x/{(x2 + 1)(x - 1)} = (-x/2)/(x2 + 1) + (1/2)/(x2 + 1) + 1/2(x - 1)

Now, ʃ [x/{(x2 + 1)(x - 1)}dx = (-1/2)ʃ [x/(x2 + 1)] + (1/2)ʃ [1/(x2 + 1)] + (1/2)ʃdx/(x - 1)

= (-1/4)ʃ [2x/(x2 + 1)] + (1/2)tan-1 x + (1/2)log|x - 1| + C

Let x2 + 1 = t

=> 2x dx = dt

So, ʃ [2x/(x2 + 1)] = ʃ dt/t

= log|t|

= log|x2 + 1|

Now, ʃ [x/{(x2 + 1)(x - 1)}dx = (-1/4) log|x2 + 1| + (1/2)tan-1 x + (1/2)log|x - 1| + C

= (1/2)log|x - 1|- (1/4) log|x2 + 1| + (1/2)tan-1 x + C

Question 8:

x/{(x - 1)2(x + 2)}

Let x/{(x - 1)2(x + 2)} = A/(x - 1) + B/(x - 1)2 + C/(x + 2)

=> x = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)2

Put x = 1, we get

1 = B(1 + 1)

=> B = 1/3

Equating the coefficients of x2 and constant terms, we get

A + C = 0

-2A + 2B + C = 0

On solving, we get

A = 2/9 and C = -2/9

So, x/{(x - 1)2(x + 2)} = 2/9(x - 1) + 1/3(x - 1)2 - 2/9(x + 2)

Now, ʃ [x/{(x - 1)2(x + 2)}]dx = (2/9)ʃ{1/(x - 1)}dx + (1/3)ʃ[1/(x - 1)2]dx – (2/9)ʃ[1/(x + 2)]dx

= (2/9) log|x  -1| + (1/3)[-1/(x - 1) – (2/9) log|x + 2| + C

= (2/9) log|(x  -1)/(x + 2)| - 1/3(x - 1) + C

Question 9:

(3x + 5)/(x3 – x2 – x + 1)

(3x + 5)/(x3 – x2 – x + 1) = (3x + 5)/{(x – 1)2(x + 1)}

Let (3x + 5)/{(x – 1)2(x + 1)} = A/(x - 1) + B/(x - 1)2 + C/(x + 1)

=> 3x + 5 = A(x - 1)(x + 1) + B/(x + 1) + C(x - 1)2

=> 3x + 5 = A(x2 - 1) + B(x + 1) + C(x2 – 2x + 1)     ………1

Put x = 1 in equation 1, we get

=> 3 * 1 + 5 = B(1 + 1)

=> 8 = 2B

=> B = 4

Equating the coefficient of x2, we get

A + C = 0

B – 2C = 3

On solving, we get

A = -1/2 and C = 1/2

So, (3x + 5)/{(x – 1)2(x + 1)} = -1/2(x - 1) + 4/(x - 1)2 + 1/2(x + 1)

Now, ʃ [(3x + 5)/{(x – 1)2(x + 1)}]dx = (-1/2)ʃ [1/(x - 1)]dx + 4 ʃ [1/(x - 1)2]dx + (1/2)ʃ[1/(x + 1)dx

= (-1/2)log|x - 1| + 4(-1)/(x - 1) + (1/2)log|x + 1| + C

= (1/2)log|(x + 1)/(x - 1)| - 4/(x - 1) + C

Question 10:

(2x - 3)/{(x2 - 1)(2x + 3)}

(2x - 3)/{(x2 - 1)(2x + 3)} = (2x - 3)/{(x + 1)(x - 1)(2x + 3)}

Let (2x - 3)/{(x + 1)(x - 1)(2x + 3)} = A/(x + 1) + B/(x - 1) + C/(2x + 3)

=> 2x - 3 = A(x - 1)(2x + 3) + B(x + 1)2x + 3) + C(x + 1)(x - 1)

=> 2x - 3 = A(2x2 + x - 3) + B(2x2 + 5x + 3) + C(x2 - 1)

=> 2x – 3 = (2A + 2B + C)x2 + (A + 5B)x + (-3A + 3B - C)

Equating the coefficient of x2, x and constant term, we get

2A + 2B + C = 2

A + 5B = 0

-3A + 3B - C = 0

On solving, we get

A = 5/2, B = -1/10 and C = -24/5

So, (2x - 3)/{(x + 1)(x - 1)(2x + 3)} = 5/2(x + 1) - 1/10(x - 1) - 24/5(2x + 3)

Now, ʃ [(2x - 3)/{(x2 - 1)(2x + 3)}]dx

= ʃ[(2x - 3)/{(x + 1)(x - 1)(2x + 3)}]dx

= (5/2)ʃ[1/(x + 1)]dx – (1/10)ʃ [1/(x - 1)]dx – (24/5)ʃ[1/(2x + 3)]dx

= (5/2) log|x + 1| – (1/10) log|x - 1| – {24/(5 * 2)} log|2x + 3| + C

= (5/2) log|x + 1| – (1/10) log|x - 1| – (12/5) log|2x + 3|+ C

Question 11:

5x/{(x + 1)(x2 - 4)}

5x/{(x + 1)(x2 - 4)} = 5x/{(x + 1)(x + 2)(x - 2)}

Let 5x/{(x + 1)(x + 2)(x - 2)} = A/(x + 1) + B/(x + 2) + C/(x - 2)

=> 5x = A(x + 2)(x - 2) + B(x + 1)(x - 2) + C(x + 1)(x + 2)   …………..1

Put x = -1, -2 and 2 in equation 1, we get

A = 5/3, B = -5/2 and C = 5/6

So, 5x/{(x + 1)(x + 2)(x - 2)} = 5/3(x + 1) - 5/2(x + 2) + 5/6(x - 2)

Now, ʃ [5x/{(x + 1)(x + 2)(x - 2)}]dx = (5/3)ʃ[1/(x + 1)dx – (5/2)ʃ[1/(x + 2)] + (5/6)ʃ[1/(x - 2)dx

= (5/3) log|x + 1| – (5/2) log|x + 2| + (5/6) log|x - 2| + C

Question 12:

(x3 + x + 1)/(x2 - 1)

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (x3 + x + 1) by (x2 – 1), we get

(x3 + x + 1)/(x2 - 1) = x + (2x + 1)/(x2 - 1) = x + (2x + 1)/{(x + 1)(x - 1)}

Let (2x + 1)/{(x + 1)(x - 1)} = A/(x + 1) + B/(x - 1)

=> 2x + 1 = A(x - 1) + B(x + 1)   ……………..1

Substituting x = 1 and −1 in equation 1, we get

A = 1/2 and B = 3/2

So, (x3 + x + 1)/(x2 - 1) = x + 1/2(x + 1) + 3/2(x - 1)

Now, ʃ [(x3 + x + 1)/(x2 - 1)]dx = ʃ x dx + (1/2)ʃ[1/(x + 1)]dx + (3/2)ʃ[1/(x - 1)]dx

= x2/2 + (1/2) log|x + 1| + (3/2) log|x - 1| + C

Question 13:

2/{(1 - x)(1 + x2)}

Let 2/{(1 - x)(1 + x2)} = A/(1 - x) + (Bx + C)/(1 + x2)

=> 2 = A(1 + x2) + (Bx + C)/(1 - x)

=> 2 = A + Ax2 + Bx – Bx2 + C – Cx

Equating the coefficient of x2, x and constant term, we get

A – B = 0

B – C = 0

A + C = 2

On solving these equations, we get

A = 1, B = 1 and C = 1

So, 2/{(1 - x)(1 + x2)} = 1/(1 - x) + (x + 1)/(1 + x2)

Now, ʃ[2/{(1 - x)(1 + x2)}]dx = ʃ dx/(1 - x) + ʃ[x/(1 + x2)]dx + ʃ[1/(1 + x2)]dx

= -ʃ dx/(x - 1) + (1/2)ʃ[2x/(1 + x2)]dx + ʃ[1/(1 + x2)]dx

= - log|x – 1| + (1/2) log|1 + x2| + tan-1 x + C

Question 14:

(3x - 1)/(x + 2)2

Let (3x - 1)/(x + 2)2 = A/(x + 2) + B/(x + 2)2

=> 3x + 1 = A(x + 2) + B

Equating the coefficient of x and constant term, we get

A = 3

2A + B = -1

=> 2 * 3 + B = -1

=> 6 + B = -1

=> B = -7

So, (3x - 1)/(x + 2)2 = 3/(x + 2) - 7/(x + 2)2

Now, (3x - 1)/(x + 2)2 = 3ʃ[1/(x + 2)]dx - 7ʃ[1/(x + 2)2]dx

= 3 log|x + 2| - 7[-1/(x + 2)] + C

= 3 log|x + 2| + 7/(x + 2) + C

Question 15:

1/(x4 - 1)

1/(x4 - 1) = 1/{(x2 - 1)(x2 + 1)} = 1/{(x + 1)(x - 1)(x2 + 1)}

Let 1/{(x + 1)(x - 1)(x2 + 1)} = A/(x + 1) + B/(x - 1) + (Cx + D)/(x2 + 1)

=> 1 = A(x - 1)(x2 + 1) + B(x + 1)(x2 + 1) + (Cx + D)(x2 - 1)

=> 1 = A(x3 + x – x2 - 1) + B(x3 + x + x2 + 1) + Cx3 + Dx2 – Cx – D

=> 1 = (A + B + C)x3 + (-A + B + D)x2 + (A + B + C)x + (-A + B - D)

Equating the coefficients of x3, x2, x and constant term, we get

A + B + C = 0

-A + B + D = 0

A + B + C = 0

-A + B – D = 1

On solving these equations, we get

A = -1/4, B = 1/4, C = 0 and D = 1/2

So, 1/(x4 - 1) = -1/4(x + 1) + 1/4(x - 1) - 1/2(x2 + 1)

Now, ʃ[1/(x4 - 1)]dx = (-1/4)ʃ[1/(x + 1)]dx + (1/4)ʃ[1/(x - 1)dx – (1/2)ʃ1/(x2 + 1)]dx

= (-1/4) log|x + 1| + (1/4) log|x - 1| – (1/2) tan-1 x + C

= (1/4) log|(x – 1)/(x + 1)| – (1/2) tan-1 x + C

Question 16:

1/{x(xn + 1)}

[Hint: multiply numerator and denominator by xn − 1 and put xn = t]

Multiplying numerator and denominator by xn − 1, we get

1/{x(xn + 1)} = xn − 1/{xn − 1 * x(xn + 1)} = xn − 1/{ xn(xn + 1)}

Let xn = t

=> n xn − 1 dx = dt

=> xn − 1 dx = dt/n

So, ʃ[1/{x(xn + 1)}]dx = ʃ[xn − 1/{xn(xn + 1)}dx = (1/n)ʃ [1/{t(t + 1)}]dt

Let 1/{t(t + 1)} = A/t + B/(t + 1)

=> 1 = A(t + 1) + Bt       …………….1

Put t = 0, -1 in equation 1, we get

A = 1 and B = -1

So, 1/{t(t + 1)} = 1/t - 1/(t + 1)

Now, ʃ[1/{x(xn + 1)}]dx = (1/n)ʃ [{1/t – 1/(t + 1)}]dt

= (1/n) [log|t| – log|t + 1|] + C

= (1/n) [log|xn| – log|xn + 1|] + C

= (1/n) [log|xn/(xn + 1)|] + C

Question 17:

cos x /{(1 - sin x)(2 – sin x)}                    [Hint: Put sin x = t]

Let sin x = t

=> cos x dx = dt

So, ʃ[cos x /{(1 - sin x)(2 – sin x)}]dx = ʃ dt/{(1 - t)(2 – t)}

Let 1/{(1 - t)(2 – t)} = A/(1 - t) + B/(2 - t)

=> 1 = A(2 - t) + B(1 - t)     ………….1

Put t = 2 and t = 1 in equation 1, we get

A = 1 and B = -1

So, 1/{(1 - t)(2 – t)} = 1/(1 - t) - 1/(2 - t)

Now, ʃ[cos x /{(1 - sin x)(2 – sin x)}]dx = ʃ [1/(1 - t) - 1/(2 - t)]dt

= -log|1 - t| + log|2 - t| + C

= log|(2 - t)/(1 - t)| + C

= log|(2 – sin x)/(1 – sin x)| + C

Question 18:

{(x2 + 1)(x2 + 2)}/{(x2 + 3)(x2 + 4)}

{(x2 + 1)(x2 + 2)}/{(x2 + 3)(x2 + 4)} = 1 - (4x2 + 10)/{(x2 + 3)(x2 + 4)}

Let (4x2 + 10)/{(x2 + 3)(x2 + 4)} = (Ax + B)/(x2 + 3) + (Cx + D)/(x2 + 4)

=> (4x2 + 10) = (Ax + B)(x2 + 4) + (Cx + D)(x2 + 3)

=> (4x2 + 10) = Ax3 + 4Ax + Bx2 + 4B + Cx3 + 3Cx + Dx2 + 3D

=> (4x2 + 10) = (A + C)x3 + (B + D)x2 + (4A + 3C)x + (4B + 3D)

Equating the coefficients of x3, x2, x and constant term, we get

A + C = 0

B + D = 0

4A + 3C = 0

4B + 3D = 10

On solving these equations, we get

A = 0, B = -2, C = 0 and D = 6

So, (4x2 + 10)/{(x2 + 3)(x2 + 4)} = -2/(x2 + 3) + 6/(x2 + 4)

Now, {(x2 + 1)(x2 + 2)}/{(x2 + 3)(x2 + 4)} = 1 – [-2/(x2 + 3) + 6/(x2 + 4)]

= 1 + 2/(x2 + 3) - 6/(x2 + 4)

Now, ʃ [{(x2 + 1)(x2 + 2)}/{(x2 + 3)(x2 + 4)}]dx = ʃ[1 + 2/(x2 + 3) - 6/(x2 + 4)]dx

= ʃ[1 + 2/{x2 + (√3)2 - 6/(x2 + 22)]dx

= x + (2/√3)tan-1 (x/√3) – (6/2) tan-1 (x/2) + C

Question 19:

2x/{(x2 + 1)(x2 + 3)}

Let x2 = t

=> 2x dx = dt

So, ʃ[2x/{(x2 + 1)(x2 + 3)}]dx = ʃ[dt/{(t + 1)(t + 3)}]   …………….1

Let 1/{(t + 1)(t + 3)} =A/(t + 1) + B/(t + 3)

=> 1 = A(t + 3) + B(t + 1)

Put t = -3 and t = -1 in equation 1, we get

A = 1/2 and B = -1/2

So, 1/{(t + 1)(t + 3)} =1/2(t + 1) - 1/2(t + 3)

Now, ʃ [2x/{(x2 + 1)(x2 + 3)}]dx = (1/2)ʃ [1/(t + 1)]dt – (1/2)ʃ [1/(t + 3)dt

= (1/2) log|t + 1| – (1/2) log|t + 3| + C

= (1/2) log|(t + 1)/(t + 3)| + C

= (1/2) log|(x2 + 1)/(x2 + 3)| + C

Question 20:

1/{x(x4 - 1)}

Multiplying numerator and denominator by x3, we get

1/{x(x4 - 1)} = x3/{x4(x4 - 1)}

Let x4 = t

=> 4x3 dx = dt

=> x3 dx = dt/4

So, ʃ[1/{x(x4 - 1)]dx = (1/4)ʃ[1/{t(t - 1)}]dt

Let 1/{t(t - 1)} = A/t + B/(t - 1)

=> 1 = A(t - 1) + Bt    ……………..1

Put t = 0 and t = 1 in equation 1, we get

A = -1 and B = 1

So, 1/{t(t - 1)} = -1/t + 1/(t - 1)

Now, ʃ[1/{x(x4 - 1)]dx = (1/4)ʃ[-1/t + 1/(t - 1)]dt

= (1/4)[-log|t| + log|t – 1|] + C

= (1/4) log|(t - 1)/t| + C

= (1/4) log|(x4 - 1)/x4| + C

Question 21:

1/(ex - 1)      [Hint: Put ex = t]

Let ex = t

=> ex dx = dt

=> dx = dt/ex

=> dx = dt/t

So, ʃ[1/(ex - 1)]dx = ʃ[1/{t(t - 1)}]dt

Let 1/{t(t - 1)} = A/t + B/(t - 1)

=> 1 = A(t - 1) + Bt    ……………..1

Put t = 0 and t = 1 in equation 1, we get

A = -1 and B = 1

So, 1/{t(t - 1)} = -1/t + 1/(t - 1)

Now,  ʃ[1/(ex - 1)]dx = ʃ [1/{t(t - 1)}dt

= ʃ [-1/t + 1/(t - 1)]dt

= -log |t| + log|t - 1| + C

= log |(t + 1)/t|+ C

= log |(ex + 1)/ex|+ C

Choose the correct answer in each of the Exercises 22 and 23.

Question 22:

ʃ [x dx/{(x - 1)(x – 2)}] equals

1. log|(x - 1)2/(x - 2)| + C B. log|(x - 2)2/(x - 1)| + C
2. log|{(x - 1)/(x - 2)}2| + C D. log|(x - 1)(x - 2)| + C

Let x/{(x - 1)(x – 2)} = A/(x - 1) + B/(x - 2)

=> x = A(x - 2) + B(x - 1)     ……………….1

Put x = -1 and x = 2 in equation 1, we get

A = -1 and B = 2

So, x/{(x - 1)(x – 2)} = -1/(x - 1) + 2/(x - 2)

Now, ʃ [x dx/{(x - 1)(x – 2)}] = ʃ [-1/(x - 1) + 2/(x - 2)]dx

= -log|x - 1| + 2log|x - 2| + C

= -log|x - 1| + log|(x – 2)2| + C

= log|{(x – 2)2/(x - 1)}| + C

Hence, the correct answer is option C.

Question 23:

ʃ [dx/{x(x2 + 1)}] equals

1. log|x| - (1/2) log(x2 + 1) + C B. log|x| + (1/2) log(x2 + 1) + C
2. -log|x| + (1/2) log(x2 + 1) + C D. (1/2)log|x| + log(x2 + 1) + C

Let 1/{x(x2 + 1)} = A/x + (Bx + C)/(x2 + 1)

Equating the coefficients of x2, x and constant term, we get

A + B = 0

C = 0

A = 1

On solving these equations, we obtain

A = 1, B = −1, and C = 0

So, 1/{x(x2 + 1)} = 1/x - x/(x2 + 1)

Now, ʃ [dx/{x(x2 + 1)}] = ʃ [1/x - x/(x2 + 1)]dx

= ʃ [1/x - 2x/2(x2 + 1)]dx

= log|x| - (1/2)log|x2 + 1| + C

Hence, the correct answer is option A.

Exercise 7.6

Integrate the functions in Exercises 1 to 22.

Question 1:

x * sin x

Let I = ʃ x * sin x dx

Taking x as first function and sin x as second function and integrating by parts, we get

I = x * ʃ sin x dx - ʃ [dx/dx * ʃ sin x dx]dx

I = x * (-cos x) - ʃ 1 * (-cos x)]dx

I = -x * cos x + ʃ cos x dx

I = -x * cos x + sin x + C

Question 2:

x * sin 3x

Let I = ʃ x * sin 3x dx

Taking x as first function and sin x as second function and integrating by parts, we get

I = x * ʃ sin 3x dx - ʃ [dx/dx * ʃ sin 3x dx]dx

I = x * (-cos 3x /3) - ʃ 1 * (-cos 3x /3)]dx

I = (-x * cos 3x)/3 + (1/3) * ʃ cos 3x dx

I = (-x * cos 3x)/3 + sin 3x /9 + C

Question 3:

x2 * ex

Let I = ʃ x2 * ex dx

Taking x2 as first function and ex as second function and integrating by parts, we get

I = x2 * ʃ ex dx - ʃ [d(x2)/dx * ʃ ex dx]dx

I = x2 * ex - ʃ [2x * ex]dx

I = x2 * ex – 2 * ʃ x * ex dx

Again integrating by parts, we get

I = x2 * ex – 2 * [x * ʃ ex dx - ʃ {d(x)/dx * ʃ ex dx}dx]

I = x2 * ex – 2 * [x * ex - ʃ ex dx]

I = x2 * ex – 2 * [x * ex - ex]

I = x2 * ex – 2x * ex + 2ex

I = ex(x2 – 2x + 2) + C

Question 4:

x * log x

Let I = ʃ x * log x dx

Taking log x as first function and x as second function and integrating by parts, we get

I = log x * ʃ x dx - ʃ [d(log x)/dx * ʃ x dx]dx

I = log x * (x2/2) - ʃ [(1/x) * (x2/2)]dx

I = log x * (x2/2) - ʃ (x/2) dx

I = (x2 * log x)/2 – x2/4 + C

Question 5:

x * log 2x

Let I = ʃ x * log 2x dx

Taking log 2x as first function and x as second function and integrating by parts, we get

I = log 2x * ʃ x dx - ʃ [d(log 2x)/dx * ʃ x dx]dx

I = log 2x * (x2/2) - ʃ [(2/2x) * (x2/2)]dx

I = log 2x * (x2/2) - ʃ (x/2) dx

I = (x2 * log 2x)/2 – x2/4 + C

Question 6:

x2 * log x

Let I = ʃ x2 * log x dx

Taking log x as first function and x2 as second function and integrating by parts, we get

I = log x * ʃ x2 dx - ʃ [d(log x)/dx * ʃ x2 dx]dx

I = log x * (x3/3) - ʃ [(1/x) * (x3/3)]dx

I = log x * (x3/3) - ʃ (x2/3) dx

I = (x3 * log x)/3 – x3/9 + C

Question 7:

x * sin-1 x

Let I = ʃ x * sin-1 x dx

Taking sin-1 x as first function and x as second function and integrating by parts, we get

I = sin-1 x * ʃ x dx - ʃ[d(sin-1 x)/dx * ʃ x dx]dx

I = sin-1 x * (x2)/2 - ʃ[1/√(1 – x2) * x2/2]dx

I = sin-1 x * (x2)/2 + (1/2) * ʃ[(-x2)/√(1 – x2)]dx

I = sin-1 x * (x2)/2 + (1/2) * ʃ[(1 - x2)/√(1 – x2) - 1/√(1 – x2)]dx

I = sin-1 x * (x2)/2 + (1/2) * ʃ[√(1 – x2) - 1/√(1 – x2)]dx

I = (x2 * sin-1 x)/2 + (1/2) * ʃ[√(1 – x2) - 1/√(1 – x2)]dx

I = (x2 * sin-1 x)/2 + (1/2) * [(x/2) * √(1 – x2) + sin-1 x /2 - sin-1 x /2] + C

I = (x2 * sin-1 x)/2 + (x/4) * √(1 – x2) + (1/4) * sin-1 x – (1/2) * sin-1 x] + C

I = (2x2 – 1)/4 * sin-1 x + (x/4) * √(1 – x2) + C

Question 8:

x * tan-1 x

Let I = ʃ x * tan-1 x dx

Taking tan-1 x as first function and x as second function and integrating by parts, we get

I = tan-1 x * ʃ x dx - ʃ[d(tan-1 x)/dx * ʃ x dx]dx

I = tan-1 x * (x2)/2 - ʃ[1/(1 + x2) * x2/2]dx

I = tan-1 x * (x2)/2 - (1/2) * ʃ[x2/(1 + x2)]dx

I = tan-1 x * (x2)/2 - (1/2) * ʃ[(x2 + 1)/(1 + x2) - 1/(1 + x2)]dx

I = tan-1 x * (x2)/2 - (1/2) * ʃ[1 - 1/√(1 – x2)]dx

I = (x2 * tan-1 x)/2 - (1/2) * [x – tan-1 x] + C

I = (x2 * tan-1 x)/2 – x/2 + (tan-1 x)/2 + C

Question 9:

x * cos-1 x

Let I = ʃ x * cos-1 x dx

Taking cos-1 x as first function and x as second function and integrating by parts, we get

I = cos-1 x * ʃ x dx - ʃ[d(cos-1 x)/dx * ʃ x dx]dx

I = cos-1 x * (x2)/2 - ʃ[-1/√(1 – x2) * (x2)/2]dx

I = (x2 * cos-1 x)/2 – (1/2) * ʃ[-1/√(1 – x2)]dx

I = (x2 * cos-1 x)/2 – (1/2) * ʃ[(1 – x2 – 1)/√(1 – x2)]dx

I = (x2 * cos-1 x)/2 – (1/2) * ʃ[√(1 – x2) + (-1)/√(1 – x2)]dx

I = (x2 * cos-1 x)/2 – (1/2) * ʃ √(1 – x2) dx - (1/2) * ꭍ [(-1)/√(1 – x2)]dx

I = (x2 * cos-1 x)/2 – (1/2) * I1 - (1/2) * cos-1 x       ……………1

Where I1 = ʃ √(1 – x2) dx

=> I1 = x * √(1 – x2) - ꭍ [d(√(1 – x2))/dx * ꭍ 1 dx]dx

=> I1 = x * √(1 – x2) - ꭍ [(-2x)/2√(1 – x2) * x]dx

=> I1 = x * √(1 – x2) - ꭍ [(-x2)/√(1 – x2)]dx

=> I1 = x * √(1 – x2) - ꭍ [(1 - x2 – 1)/√(1 – x2)]dx

=> I1 = x * √(1 – x2) - ꭍ [√(1 – x2) dx + ꭍ -dx/√(1 – x2)]

=> I1 = x * √(1 – x2) - [I1 + cos-1 x]

=> I1 = x * √(1 – x2) - I1 - cos-1 x

=> 2I1 = x * √(1 – x2) - cos-1 x

=> I1 = (x/2) * √(1 – x2) – (cos-1 x)/2

From equation 1, we get

I = (x2 * cos-1 x)/2 – (1/2) * [(x/2) * √(1 – x2) – (cos-1 x)/2] - (1/2) * cos-1 x

I = (2x2 – 1)/4 * cos-1 x – (x/4) * √(1 – x2) + C

Question 10:

(sin-1 x)2

Let I = ꭍ (sin-1 x)2 * 1 dx

Taking (sin-1 x)2 as first function and 1 as second function and integrating by parts, we get

I = (sin-1 x)2 * ꭍ 1 dx - ꭍ[d{(sin-1 x)2}/dx * ꭍ 1 dx]dx

I = (sin-1 x)2 * x - ꭍ[{2 * sin-1 x}/√(1 – x2) * x]dx

I = x * (sin-1 x)2 + ꭍ[sin-1 x * {(-2x)/√(1 – x2)]dx

I = x * (sin-1 x)2 + [sin-1 x * ꭍ (-2x)/√(1 – x2)dx – {ꭍ d(sin-1 x)/dx * ꭍ (-2x)/√(1 – x2)dx}dx]

I = x * (sin-1 x)2 + [sin-1 x * 2√(1 – x2) – {ꭍ 1/√(1 – x2) * {2√(1 – x2)}dx]

I = x * (sin-1 x)2 + sin-1 x * 2√(1 – x2) – ꭍ 2 dx

I = x * (sin-1 x)2 + 2√(1 – x2) * sin-1 x – 2x  +C

Question 11:

(x * cos-1 x)/√(1 – x2)

Let I = ꭍ [(x * cos-1 x)/√(1 – x2)]dx

=> I = (-1/2) * ꭍ [(-2x)/√(1 – x2)] * cos-1 x dx

Taking cos-1 x as first function and (-2x)/√(1 – x2) as second function and integrating by parts,

we get

=> I = (-1/2) * [cos-1 x * ꭍ {(-2x)/√(1 – x2)} - ꭍ {d(cos-1 x)/dx * ꭍ (-2x)/√(1 – x2)dx}dx]

=> I = (-1/2) * [cos-1 x * 2√(1 – x2) - ꭍ {(-1)/ (1 – x2) * 2√(1 – x2)}dx]

=> I = (-1/2) * [2√(1 – x2) * cos-1 x + ꭍ 2 dx]

=> I = (-1/2) * [2√(1 – x2) * cos-1 x + 2x] + C

=> I = -[√(1 – x2) * cos-1 x + x] + C

Question 12:

x * sec2 x

Let I = ʃ x * sec2 x dx

Taking x as first function and sec2 x as second function and integrating by parts, we get

I = x * ʃ sec2 x dx - ʃ[d(x)/dx * ʃ sec2 x dx]dx

I = x * tan x - ʃ 1 * tan x dx

I = x * tan x - ʃ tan x dx

I = x * tan x + log|cos x| + C

Question 13:

tan-1 x

Let I = ʃ 1 * tan-1 x dx

Taking tan-1 x as first function and 1 as second function and integrating by parts, we get

I = tan-1 x * ʃ 1 dx - ʃ[d(tan-1 x)/dx * ʃ 1 dx]dx

I = tan-1 x * x - ʃ[1/(1 + x2) * x]dx

I = x * tan-1 x - ʃ[x/(1 + x2)]dx

I = x * tan-1 x – (1/2) * ʃ[2x/(1 + x2)]dx

I = x * tan-1 x – (1/2) * log|1 + x2| + C

I = x * tan-1 x – (1/2) * log(1 + x2) + C

Question 14:

x * (log x)2

Let I = x * (log x)2

Taking tan-1 x as first function and 1 as second function and integrating by parts, we get

I = (log x)2 * ʃ x dx - ʃ[d({(log x)2}/dx * ʃ x dx]dx

I = (log x)2 * x2/2 - ʃ[2 * log x * (1/x) * x2/2]dx

I = (log x)2 * x2/2 - ʃ x * log x dx

Again, integrating by parts, we get

I = (log x)2 * x2/2 – [(log x) * ʃ x dx - ʃ{d(log x)/dx * ʃ x dx}dx]

I = (log x)2 * x2/2 – [(log x) * (x2/2) - ʃ{(1/x) * (x2/2)}dx]

I = (log x)2 * x2/2 – [(log x) * (x2/2) – (1/2) * ʃ x dx]

I = (log x)2 * x2/2 – [(log x) * (x2/2) – (1/2) * (x2/2)] + C

I = (log x)2 * x2/2 – (log x) * (x2/2) + x2/4 + C

Question 15:

(x2 + 1) * log x

Let I = ꭍ (x2 + 1) * log x dx

=> I = ꭍ x2 * log x dx + ꭍ log x dx

=> I = I1 + I2    ……………..1

Where I1 = ꭍ x2 * log x dx and I2 = ꭍ log x dx

Now, consider I1 = ꭍ x2 * log x dx

Taking log x as first function and x2 as second function and integrating by parts, we get

I1 = log x * ꭍ x2 dx - ꭍ[d(log x)/dx * ꭍ x2 dx]dx

I1 = log x * x3/3 - ꭍ[(1/x) * (x3/3)]dx

I1 = (x3 * log x)/3 – (1/3) * ꭍ x2 dx

I1 = (x3 * log x)/3 – (1/3) * x3/3 + C1

I1 = (x3 * log x)/3 – x3/9 + C1     …………..2

Again consider I2 = ꭍ log x dx

=> I2 = ꭍ log x * 1 dx

Taking log x as first function and 1 as second function and integrating by parts, we get

=> I2 = log x * ꭍ 1 dx - ꭍ[d(log x)/dx * ꭍ 1 dx]dx

=> I2 = log x * x - ꭍ[(1/x) * x]dx

=> I2 = log x * x - ꭍ 1 dx

=> I2 = log x * x – x + C2    ………….3

From equation 1, 2 and 3, we get

=> I = (x3 * log x)/3 – x3/9 + C1 + log x * x – x + C2

=> I = (x3 * log x)/3 – x3/9 + C1 + log x * x – x + (C1 + C2)

=> I = (x3/3 + x)log x – x3/9 – x + C

Question 16:

ex(sin x + cos x)

Let I = ʃ ex(sin x + cos x) dx

Again let f(x) = sin x

=> f’(x) = cos x

So, I = ʃ ex{f(x) + f’(x)} dx

It is known that ʃ ex{f(x) + f’(x)} dx = ex f(x) + C

So, I = ex sin x + C

Question 17:

xex/(1 + x)2

Let I = ꭍ[xex/(1 + x)2]dx

=> I = ꭍ[ex{x/(1 + x)2}]dx

=> I = ꭍ[ex{(1 + x - 1)/(1 + x)2}]dx

=> I = ꭍ[ex{(1 + x)/(1 + x)2 - 1/(1 + x)2}]dx

=> I = ꭍ[ex{1/(1 + x) - 1/(1 + x)2}]dx

Let f(x) = 1/(x + 1)

So, f’(x) = -1/(1 + x)2

Now, ꭍ[xex/(1 + x)2]dx = ʃ ex{f(x) + f’(x)} dx

It is known that ʃ ex{f(x) + f’(x)} dx = ex f(x) + C

So, ꭍ[xex/(1 + x)2]dx = ex/(1 + x) + C

Question 18:

ex(1 + sin x)/(1 + cos x)

Given, ex(1 + sin x)/(1 + cos x) = ex(sin2 x/2 + cos2 x/2 + 2 * sin x/2 * cos x/2)/(2 cos2 x/2)

= ex(sin x/2 + cos x/2)2/(2 cos2 x/2)

= (ex/2) * {(sin x/2 + cos x/2)/(cos x/2)}2

= (ex/2) * (tan x/2 + 1)2

= (ex/2) * (1 + tan2 x/2 + 2 * tan x/2)

= (ex/2) * (sec2 x/2 + 2 * tan x/2)

= ex * {(1/2) * (sec2 x/2) + tan x/2}

Let tan x/2 = f(x)

So, f’(x) = (1/2) * (sec2 x/2)

It is known that ʃ ex{f(x) + f’(x)} dx = ex f(x) + C

So, ꭍ[ ex(1 + sin x)/(1 + cos x)]dx = ex * tan x/2 + C

Question 19:

ex(1/x – 1/x2)

Let I = ʃ ex(1/x – 1/x2) dx

Again let 1/x = f(x)

=> f’(x) = -1/x2

It is known that ʃ ex{f(x) + f’(x)} dx = ex f(x) + C

So, I = ex/x + C

Question 20:

{(x - 3)ex}/(x - 1)3

ꭍ [{(x - 3)ex}/(x - 1)3]dx = ꭍ [ex * (x - 3)/(x - 1)3]dx

= ꭍ [ex * (x – 1 - 2)/(x - 1)3]dx

= ꭍ [ex * {(x – 1)/(x - 1)3 - 2/(x - 1)3]dx

= ꭍ [ex * {1/(x - 1)2 - 2/(x - 1)3]dx

Let f(x) = 1/(x - 1)2

So, f’(x) = -2/(x - 1)3

It is known that ʃ ex{f(x) + f’(x)} dx = ex f(x) + C

So, ꭍ [{(x - 3)ex}/(x - 1)3]dx = ex/(x - 1)2 + C

Question 21:

e2x * sin x

Let I = e2x * sin x

Integrating by parts, we get

I = sin x * ꭍ e2x dx - ꭍ[d(sin x)/dx * ꭍ e2x dx]dx

=> I = sin x * e2x/2 - ꭍ[cos x * e2x/2]dx

=> I = (e2x * sin x)/2 – (1/2) * ꭍ e2x * cos x dx

Again integrating by parts, we get

I = (e2x * sin x)/2 – (1/2) * [cos x * ꭍ e2x dx - ꭍ{d(cos x)/dx * ꭍ e2x dx}dx]

=> I = (e2x * sin x)/2 – (1/2) * [cos x * e2x/2 - ꭍ{(-sin x) * e2x/2}dx]

=> I = (e2x * sin x)/2 – (1/2) * [cos x * e2x/2 + (1/2) * ꭍ{sin x * e2x}dx]

=> I = (e2x * sin x)/2 – (e2x * cos x)/2 – I/4

=> I + I/4 = (e2x * sin x)/2 – (e2x * cos x)/4

=> 5I/4 = (e2x * sin x)/2 – (e2x * cos x)/2

=> I = 4[(e2x * sin x)/2 – (e2x * cos x)/4]/5 + C

=> I = e2x[2 * sin x - cos x)]/5 + C

Question 22:

sin-1{2x/(1 + x2)}

Let x = tan θ

=> dx = sec2 θ dθ

So, sin-1{2x/(1 + x2)} = sin-1{(2 * tan θ)/(1 + tan2 θ)}

= sin-1(sin 2θ)

= 2θ

Now, ꭍ sin-1{2x/(1 + x2)} dx = ꭍ[2θ * sec2 θ]dθ

= 2 * ꭍ[θ * sec2 θ]dθ

Integrating by parts, we get

2 * ꭍ[θ * sec2 θ]dθ = 2[θ * ꭍ sec2 θ dθ - ꭍ{d(θ)/dθ * ꭍ sec2 θ dθ}dθ]

=> 2 * ꭍ[θ * sec2 θ]dθ = 2[θ * tan θ - ꭍ tan θ dθ]

=> 2 * ꭍ[θ * sec2 θ]dθ = 2[θ * tan θ + log|cos θ|] + C

=> 2 * ꭍ[θ * sec2 θ]dθ = 2x * tan-1 x + 2 * log|1/(1 + x2)| + C

=> 2 * ꭍ[θ * sec2 θ]dθ = 2x * tan-1 x + 2 * log(1 + x2)-1/2 + C

=> 2 * ꭍ[θ * sec2 θ]dθ = 2x * tan-1 x + 2 * (-1/2) * log(1 + x2) + C

=> 2 * ꭍ[θ * sec2 θ]dθ = 2x * tan-1 x - log(1 + x2) + C

Hence, ꭍ sin-1{2x/(1 + x2)} dx = 2x * tan-1 x - log(1 + x2) + C

Choose the correct answer in Exercises 23 and 24.

Question 23:

ʃ x2 * ex3 dx equals

(A) ex3/3 + C                      (B) ex2/3 + C                        (C) ex3/2 + C                         (D) ex2/2 + C

Let I = ʃ x2 * ex3 dx

Again let x3 = t

=> 3x2 dx = dt

So, I = (1/3) * ʃ et dt

=> I = (1/3) * et + C

=> I = (1/3) * ex3 + C

Hence, the correct answer is option A.

Question 24:

ʃex * sec x * (1 + tan x)dx equals

(A) ex * cos x + C            (B) ex * sec x + C            (C) ex * sin x + C              (D) ex * tan x + C

Let I = ʃex * sec x * (1 + tan x)dx

Again let sec x = f(x)

=> secx * tan x = f’(x)

It is known that ʃ ex{f(x) + f’(x)} dx = ex f(x) + C

So, I = ex * sec x + C

Hence, the correct answer is option B.

Exercise 7.7

Integrate the functions in Exercises 1 to 9.

Question 1:

√(4 – x2)

Let I = ʃ √(4 – x2)dx

We know that ʃ √(a2 – x2) dx = (x/2) * √(a2 – x2) + (a2/2) * sin-1(x/a) + C

So, I = (x/2) * √(4 – x2) + (4/2) * sin-1(x/2) + C

=> I = (x/2) * √(4 – x2) + 2 * sin-1(x/2) + C

Question 2:

√(1 – 4x2)

Let I = √(1 – 4x2) = √{12 – (2x)2}

Again let 2x = t

=> 2 dx = dt

So, I = (1/2) √(12 – t2)

We know that ʃ √(a2 – x2) dx = (x/2) * √(a2 – x2) + (a2/2) * sin-1(x/a) + C

So, I = (1/2)[(t/2) * √(1 – t2) + (1/2) * sin-1 t] + C

=> I = (t/4) * √(1 – t2) + (1/4) * sin-1 t + C

=> I = (2x/4) * √(1 – 4x2) + (1/4) * sin-1 2x + C

=> I = (x/2) * √(1 – 4x2) + (1/4) * sin-1 2x + C

Question 3:

√(x2 + 4x + 6)

Let I = ʃ √(x2 + 4x + 6) dx

=> I = ʃ √(x2 + 4x + 4 + 2) dx

=> I = ʃ √{(x2 + 4x + 4) + 2} dx

=> I = ʃ √{(x + 2)2 + (√2)2} dx

We know that ʃ √(x2 + a2) dx = (x/2) * √(x2 + a2) + (a2/2) * log|x + √(x2 + a2)| + C

So, I = {(x + 2)/2} * √(x2 + 4x + 6) + (2/2) * log|(x + 2) + √(x2 + 4x + 6)| + C

=> I = {(x + 2)/2} * √(x2 + 4x + 6) + log|(x + 2) + √(x2 + 4x + 6)| + C

Question 4:

√(x2 + 4x + 1)

Let I = ʃ √(x2 + 4x + 1) dx

=> I = ʃ √(x2 + 4x + 4 - 3) dx

=> I = ʃ √{(x2 + 4x + 4) - 3} dx

=> I = ʃ √{(x + 2)2 - (√3)2} dx

We know that ʃ √(x2 - a2) dx = (x/2) * √(x2 - a2) - (a2/2) * log|x + √(x2 - a2)| + C

So, I = {(x + 2)/2} * √(x2 + 4x + 1) - (3/2) * log|(x + 2) + √(x2 + 4x + 1)| + C

Question 5:

√(1 – 4x - x2)

Let I = ʃ √(1 – 4x - x2) dx

=> I = ʃ √{1 – (x2 + 4x + 4 – 4)} dx

=> I = ʃ √{1 + 4 – (x + 2)2} dx

=> I = ʃ √{(√5)2 – (x + 2)2} dx

We know that ʃ √(a2 - x2) dx = (x/2) * √(a2 - x2) - (a2/2) * sin-1 (x/a) + C

So, I = [(x + 2)/2] * √(1 – 4x - x2) + (5/2) * sin-1{(x + 2)/√5} + C

Question 6:

√(x2 + 4x - 5)

Let I = ʃ √(x2 + 4x - 5) dx

=> I = ʃ √(x2 + 4x + 4 - 9) dx

=> I = ʃ √{(x2 + 4x + 4) - 9} dx

=> I = ʃ √{(x + 2)2 - (3)2} dx

We know that ʃ √(x2 - a2) dx = (x/2) * √(x2 - a2) - (a2/2) * log|x + √(x2 - a2)| + C

So, I = {(x + 2)/2} * √(x2 + 4x - 5) - (9/2) * log|(x + 2) + √(x2 + 4x - 5)| + C

Question 7:

√(1 + 3x - x2)

Let I = ʃ √(1 + 3x - x2) dx

=> I = ʃ √{1 – (x2 - 3x + 9/4 – 9/4)} dx

=> I = ʃ √{1 + 9/4 – (x – 3/2)2} dx

=> I = ʃ √{(√13/2)2 – (x – 3/2)2} dx

We know that ʃ √(a2 - x2) dx = (x/2) * √(a2 - x2) - (a2/2) * sin-1 (x/a) + C

So, I = [(x – 3/2)/2] * √(1 + 3x - x2) + {13/(4 *2)} * sin-1 {(x – 3/2)/(√13/2)} + C

=> I = [(2x – 3)/4] * √(1 + 3x - x2) + (13/8) * sin-1 {(2x – 3)/√13} + C

Question 8:

√(x2 + 3x)

Let I = ʃ √(x2 + 3x) dx

=> I = ʃ √(x2 + 3x + 9/4 – 9/4) dx

=> I = ʃ √{(x + 3/2)2 - (3/2)2} dx

We know that ʃ √(x2 - a2) dx = (x/2) * √(x2 - a2) - (a2/2) * log|x + √(x2 - a2)| + C

So, I = {(x + 3/2)/2} * √(x2 + 3x) – {(9/4)/2} * log|(x + 3/2) + √(x2 + 3x)| + C

=> I = {(2x + 3)/2} * √(x2 + 3x) – (9/8) * log|(x + 3/2) + √(x2 + 3x)| + C

Question 9:

√(1 + x2/9)

Let I = ʃ √(1 + x2/9) dx = (1/3) ʃ √(9 + x2) dx

We know that ʃ √(x2 + a2) dx = (x/2) * √(x2 + a2) + (a2/2) * log|x + √(x2 + a2)| + C

So, I = (1/3)[(x/2) * √(x2 + 9) + (9/2) * log|x + √(x2 + 9)|] + C

=> I = (x/6) * √(x2 + 9) + (3/2) * log|x + √(x2 + 9)|] + C

Choose the correct answer in Exercises 10 to 11.

Question 10:

ʃ √(1 + x2)dx is equal to

1. (x/2) * √(1 + x2) + (1/2) * log|x + √(1 + x2)| + C
2. (2/3) * √(1 + x2)2/3 + C
3. (2x/3) * √(1 + x2)3/2 + C
4. (x2/2) * √(1 + x2) + (x2/2) * log|x + √(1 + x2)| + C

We know that ʃ √(a2 + x2) dx = (x/2) * √(a2 + x2) + (a2/2) * log|x + √(a2 + x2)| + C

So, ʃ √(1 + x2) dx = (x/2) * √(1 + x2) + (1/2) * log|x + √(1 + x2)| + C

Hence, the correct answer is option A.

Question 11:

ʃ √(x2 - 8x + 7) dx is equal to

1. {(x - 4)/2} * √(x2 - 8x + 7) + 9 * log|(x - 4) + √(x2 - 8x + 7)| + C
2. {(x + 4)/2} * √(x2 - 8x + 7) + 9 * log|(x + 4) + √(x2 - 8x + 7)| + C
3. {(x - 4)/2} * √(x2 - 8x + 7) - 3√2 * log|(x - 4) + √(x2 - 8x + 7)| + C
4. {(x - 4)/2} * √(x2 - 8x + 7) – (9/2) * log|(x - 4) + √(x2 - 8x + 7)| + C

Let I = ʃ √(x2 - 8x + 7) dx

=> I = ʃ √(x2 - 8x + 16 - 9) dx

=> I = ʃ √{(x2 - 8x + 16) - 9} dx

=> I = ʃ √{(x - 4)2 - (3)2} dx

We know that ʃ √(x2 - a2) dx = (x/2) * √(x2 - a2) - (a2/2) * log|x + √(x2 - a2)| + C

So, I = {(x - 4)/2} * √(x2 - 8x + 7) - (9/2) * log|(x - 4) + √(x2 - 8x + 7)| + C

Hence, the correct answer is option D.

Exercise 7.8

Evaluate the following definite integrals as limit of sums.

Question 1:

ʃab x dx

We know that

ʃab f(x) dx = (b - a)limn-> (1/n)[f(a) + f(a + h) + …………..+ f{a + (n - 1)h}], where h = (b - a)/n

Here a = a, b = b and f(x) = x

So, ʃab x dx = (b - a)limn-> (1/n)[a + (a + h) + (a + 2h) +…………..+ a + (n - 1)h]

= (b - a)limn-> (1/n)[(a + a + a +……..n times) + {h + 2h + 3h +………+ (n - 1)h}]

= (b - a)limn-> (1/n)[na + h{1 + 2 + 3 +………+ (n - 1)]

= (b - a)limn-> (1/n)[na + h{n(n - 1)}/2]

= (b - a)limn-> (1/n)[na + {n(n - 1)h}/2]

= (b - a)limn-> (n/n)[a + {(n - 1)h}/2]

= (b - a)limn-> [a + {(n - 1)h}/2]

= (b - a)limn-> [a + {(n - 1)(b - a)}/2n]                           [Since h = (b - a)/n]

= (b - a)limn-> [a + {(1 – 1/n)(b - a)}/2]

= (b - a)[a + {(1 – 1/∞)(b - a)}/2]

= (b - a)[a + {(1 – 0)(b - a)}/2]

= (b - a)[a + (b - a)/2]

= (b - a)[2a + (b - a)]/2

= (b - a)(b + a)/2

= (b2 – a2) /2

Question 2:

ʃ05 (x + 1) dx

We know that

ʃab f(x) dx = (b - a)limn-> (1/n)[f(a) + f(a + h) + …………..+ f{a + (n - 1)h}], where h = (b - a)/n

Here a = 0, b = 5 and f(x) = x + 1

So, h = (5 - 0)/n = 5/n

Now, ʃ05 (x + 1) dx = (5 - 0)limn-> (1/n)[f(0) + f(5/h) + …………..+ f{5(n - 1)/n}]

= 5 * limn-> (1/n)[1 + (5/n + 1) + …………..+ {1 + 5(n - 1)/n}]

= 5 * limn-> (1/n)[(1 + 1 + 1 + …….n times) +

{5/n + 2 * 5/n + …………..+ 5(n - 1)/n}]

= 5 * limn->(1/n)[(1 + 1 + …….n times) + (5/n){1 + 2 + 3 +…………..+ (n - 1)}]

= 5 * limn-> (1/n)[n + (5/n) * n(n - 1)/2]

= 5 * limn-> (1/n)[n + 5(n - 1)/2]

= 5 * limn-> (n/n)[1 + 5(1 – 1/n)/2]

= 5 * [1 + 5(1 – 1/∞)/2]

= 5 * [1 + 5(1 – 0)/2]

= 5 * [1 + 5/2]

= 5 * (7/2)

= 35/2

Question 3:

ʃ23 x2 dx

We know that

ʃab f(x) dx = (b - a)limn-> (1/n)[f(a) + f(a + h) + …………..+ f{a + (n - 1)h}], where h = (b - a)/n

Here a = 2, b = 3 and f(x) = x2

So, h = (3 - 2)/n = 1/n

Now, ʃ23 x2 dx = (3 - 2)limn-> (1/n)[f(2) + f(2 + 1/n) + …………..+ f{2 + (n - 1)/n}]

= limn-> (1/n)[22 + (2 + 1/n)2 + …………..+ {2 + (n - 1)/n}2]

= limn-> (1/n)[22 + {(22 + (1/n)2 2 * 2 * (1/n)} + …………..+

{22 + (n - 1)2/n2 + 2 * 2 * (n - 1)/n]

= limn-> (1/n)[(22 + 22 +……..n times) + {(1/n)2 + (2/n)2 +……..+ {(n - 1)/n}2

+ 2 * 2 * {1/n + 2/n  +3/n +………+ (n - 1)/n}]

= limn-> (1/n)[4n + (1/n2){12 + 22 +……..+ (n - 1)2 } + (4/n){1 + 2 + 3 +………+ (n - 1)}]

= limn-> (1/n)[4n + (1/n2){n(n - 1)(2n - 1)}/6 + (4/n){n(n - 1)}/2]

= limn-> (1/n)[4n + {n(1 – 1/n)(2 – 1/n)}/6 + {4(n - 1)}/2]

= limn-> (n/n)[4 + (1/6)(1 – 1/n)(2 – 1/n) + 2(1 – 1/n)]

= 4 + (1/6)(1 – 1/∞)(2 – 1/∞) + 2(1 – 1/∞)

= 4 + (1/6)(1 – 0)(2 – 0) + 2(1 – 0)

= 4 + 2/6 + 2

= 6 + 1/3

= 19/3

Question 4:

ʃ14 (x2 – x) dx

Let I = ʃ14 (x2 – x) dx

=> I = ʃ14 x2 dx – ʃ14 x dx

=> I = I1 - I2 ……….1

Where I1 = ʃ14 x2 dx and I2 = ʃ14 x dx

We know that

ʃab f(x) dx = (b - a)limn-> (1/n)[f(a) + f(a + h) + …………..+ f{a + (n - 1)h}], where h = (b - a)/n

For I1 = ʃ14 x2 dx

Here a = 1, b = 4 and f(x) = x2

So, h = (4 - 1)/n = 3/n

Now, I1 = ʃ14 x2 dx = (4 - 1)limn-> (1/n)[f(1) + f(1 + h) + …………..+ f{1 + (n - 1)h}]

= 3 * limn-> (1/n)[12 + (1 + 3/n)2 + …………..+ {1 + 3(n - 1)/n}2]

= 3 * limn-> (1/n)[12 + {12 + (3/n)2 + 2 * (3/n)2 + 2 * (3/n)2} + …………..

+ {12 + {3(n - 1)/n}2 + {2 * 3(n - 1)}/n]

= 3 * limn-> (1/n)[(12 + 12 +……..n times) + (3/n)2{12 + 22 +……..+ (n - 1)}2

+ 2 * (3/n) * {1 + 2 + 3 +………+ (n - 1)}]

= 3 * limn-> (1/n)[n + (9/n2){n(n - 1)(2n - 1)}/6 + (6/n){n(n - 1)}/2]

= 3 * limn-> (1/n)[1 + (9n/6)(1 – 1/n)(2 – 1/n) + (6n - 6)/2]

= 3 * limn-> [1 + (9/6)(1 – 1/n)(2 – 1/n) + 3 – 3/n]

= 3[1 + 3 + 3]

= 3 * 7

= 21

=> I1 = 21   ……………2

For I2 = ʃ14 x dx

Here a = 1, b = 4 and f(x) = x

So, h = (4 - 1)/n = 3/n

So, ʃab x dx = (4 - 1)limn-> (1/n)[f(1) + f(1 + h) + …………..+ f{a + (n - 1)h}]

= 3 * limn-> (1/n)[1 + (1 + h) + …………+ {1 + (n - 1)h}]

= 3 * limn-> (1/n)[ 1 + (1 + 3/n) + …………+ {1 + 3(n - 1)/n}]

= 3 * limn-> (1/n)[(1 + 1 + …….n times) + (3/n){1 + 2 + 3 + ………..+ (n - 1)}]

= 3 * limn-> (1/n)[n + (3/n){n(n - 1)}/2]

= 3 * limn-> [1 + (3/2)(1 – 1/n)]

= 3 * [1 + (3/2)(1 – 1/∞)]

= 3 * (1 + 3/2)

= 3 * (5/2)

= 15/2

=> I2 = 15/2   ……………3

From equation 1, 2 and 3, we get

I = 21 - 15/2

=> I = 27/2

=> ʃ14 (x2 – x) dx = 27/2

Question 5:

ʃ-11 ex dx

We know that

ʃab f(x) dx = (b - a)limn-> (1/n)[f(a) + f(a + h) + …………..+ f{a + (n - 1)h}], where h = (b - a)/n

Here a = -1, b = 1 and f(x) = ex

So, h = (1 + 1)/n = 2/n

Now, ʃ23 ex dx = (1 + 1)limn-> (1/n)[f(-1) + f(-1 + 1/n) + f(-1 + 2 * 2/n) +……+ f{-1 + 2(n - 1)/n}]

= 2 * limn-> (1/n)[ex + e(-1 + 2/n) + e(-1 + 2 * 2/n)………..……+ e{-1 + 2(n - 1)/n}]

= 2 * limn-> (1/n)[e-1{ 1 + e 2/n + e4/n………..……+ e2(n - 1)/n}}]

= 2 * limn-> (e-1/n)[1 + e 2/n + e4/n………..……+ e2(n - 1)/n}]

= 2 * limn-> (e-1/n)[e2n/n - 1 /e2/n - 1]

= 2 * limn-> (e-1/n)[(e2 -1) /e2/n - 1]

= e-1 * 2 * limn-> (1/n)[(e2 -1) /e2/n - 1]

= [e-1 * 2(e2 -1)]/[lim2/n->0 {(e2/n - 1])/(2/n)} * 2]

= [e-1 * 2(e2 -1)]/2                                                      [Since limh->0 (eh - 1)/h = 1]

= (e2 -1)/e

= (e – 1/e)

Question 6:

ʃ04 (x + e2x) dx

We know that

ʃab f(x) dx = (b - a)limn-> (1/n)[f(a) + f(a + h) + …………..+ f{a + (n - 1)h}], where h = (b - a)/n

Here a = 0, b = 4 and f(x) = x + ex

So, h = (4 - 0)/n = 4/n

Now, ʃ05 (x + 1) dx = (4 - 0)limn-> (1/n)[f(0) + f(h) + f(2h) + …………..+ f{(n - 1)h}]

= 4 * limn-> (1/n)[(0 + e0) + (h + e2h) + (2h + e2*2h) +………+ {(n - 1)h + e2(n-1)h}]

= 4 * limn-> (1/n)[1 + (h + e2h) + (2h + e4h) +………+ {(n - 1)h + e2(n-1)h}

= 4 * limn->(1/n)[{h + 2h + 3h + …….(n - 1)h} + {1 + e2h + e4h +……..+ e2(n - 1)h}

= 4 * limn-> (1/n)[h{1+ 2 + 3 + …….(n - 1)} + (e2hn - 1)/(e2h - 1)]

= 4 * limn-> (1/n)[h{n(n - 1)/2} + (e2hn - 1)/(e2h - 1)]

= 4 * limn-> (1/n)[(4/n){n(n - 1)/2} + (e8 - 1)/(e8/n - 1)]

= 4 * 2 + 4 * limn->[(e8 - 1)/{(e8/n - 1)/(8/n) * 8}]

= 8 + {4(e8 - 1)}/8                                                 [Since limx->0 (ex - 1)/x = 1]

= 8 + (e8 - 1)/2

= (15 + e8)/2

Exercise 7.9

Evaluate the definite integrals in Exercises 1 to 20.

Question 1:

-11 (x + 1) dx

Let I = ꭍ-11 (x + 1) dx

Now, ꭍ (x + 1) dx = x2/2 + x = F(x)

By second fundamental theorem of calculus, we get

I = F(1) - F(-1)

= (1/2 + 1) – (1/2 - 1)

= 1/2 + 1 – 1/2 + 1

= 2

So, ꭍ-11 (x + 1) dx = 2

Question 2:

21 dx/x

Let I = ꭍ21 dx/x

Now, ꭍ dx/x = log|x| = F(x)

By second fundamental theorem of calculus, we get

I = F(3) - F(2)

= log|3| - log|2|

= log 3 – log 2

= log(3/2)

So, ꭍ21 dx/x = log(3/2)

Question 3:

12 (4x3 – 5x2 + 6x + 9) dx

Let I = ꭍ12 (4x3 – 5x2 + 6x + 9) dx

Now, ꭍ12 (4x3 – 5x2 + 6x + 9) dx = 4(x4/4) – 5(x3/3) + 6(x2/2) + 9x

= x4 – 5x3/3 + 3x2/2 + 9x

= F(x)

By second fundamental theorem of calculus, we get

I = F(2) - F(1)

= [24 – (5 * 23)/3 + (3 * 22)/2 + 9 * 2] – [14 – (5 * 13)/3 + (3 * 12)/2 + 9 * 1]

= (16 – 40/3 + 12 + 18) – (1 – 5/3 + 3 + 9)

= 16 – 40/3 + 12 + 18) – 1 + 5/3 - 3 – 9

= 33 – 35/3

= (33 * 3 - 35)/3

= (99 - 35)/3

= 64/3

So, ꭍ12 (4x3 – 5x2 + 6x + 9) dx = 64/3

Question 4:

0π/4 sin 2x dx

Let I = ꭍ0π/4 sin 2x dx

So, ꭍ sin 2x dx = (-cos 2x)/2 = F(x)

By second fundamental theorem of calculus, we get

I = F(π/4) – F(0)

= (-1/2) * [cos 2π/4 – cos 0]

= (-1/2) * [cos π/2 – cos 0]

= (-1/2) * [0 – 1]

= 1/2

Question 5:

0π/2 cos 2x dx

Let I = ꭍ0π/2 cos 2x dx

So, ꭍ cos 2x dx = (sin 2x)/2 = F(x)

By second fundamental theorem of calculus, we get

I = F(π/2) – F(0)

= (1/2) * [sin 2π/2 – sin 0]

= (1/2) * [sin π – sin 0]

= (1/2) * [0 – 0]

= 0

Question 6:

45 ex dx

Let I = ꭍ45 ex dx

So, ꭍ ex dx = ex = F(x)

By second fundamental theorem of calculus, we get

I = F(5) – F(4)

= e5 – e4

= e4(e – 1)

Question 7:

0π/4 tan x dx

Let I = ꭍ0π/4 tan x dx

So, ꭍ tan x dx = -log|cos x| = F(x)

By second fundamental theorem of calculus, we get

I = F(π/4) – F(0)

= -log|cos π/4| + log|cos 0|

= -log|1/√2| + log|1|

= -log|1/√2|

= -log(2)-1/2

= (1/2) * log 2

Question 8:

π/6π/4 cosec x dx

Let I = ꭍπ/6π/4 cosec x dx

So, ꭍ cosec x dx = log|cosec x – cot x| = F(x)

By second fundamental theorem of calculus, we get

I = F(π/4) – F(π/6)

= log|cosec π/4 - cot π/4| - log|cosec π/6 - cot π/6|

= log|√2 - 1| - log|2 - √3|

= log{(√2 – 1)/( 2 - √3)|

Question 9:

01 dx/√(1 – x2)

Let I = ꭍ01 dx/√(1 – x2)

So, ꭍ dx/√(1 – x2) = sin-1 x = F(x)

By second fundamental theorem of calculus, we get

I = F(1) – F(0)

= sin-1 1 - sin-1 0

= π/2 – 0

= π/2

Question 10:

01 dx/(1 + x2)

Let I = ꭍ01 dx/(1 + x2)

So, ꭍ dx/(1 + x2) = tan-1 x = F(x)

By second fundamental theorem of calculus, we get

I = F(1) – F(0)

= tan-1 1 - tan-1 0

= π/4 – 0

= π/4

Question 11:

23 dx/(x2 – 1)

Let I = ꭍ23 dx/(x2 – 1)

So, ꭍ dx/(x2 – 1) = (1/2) * log|(x - 1)/(x + 1)| = F(x)

By second fundamental theorem of calculus, we get

I = F(3) – F(2)

= (1/2) * [log|(3 - 1)/(3 + 1)| - log|(2 - 1)/(2 + 1)|]

= (1/2) * [log|2/4| - log|1/3|]

= (1/2) * [log|1/2| - log|1/3|]

= (1/2) * [log (1/2)/(1/3)]

= (1/2) * log (3/2)

Question 12:

0π/2 cos2 x dx

Let I = ꭍ0π/2 cos2 x dx

So, ꭍ cos2 x dx = ꭍ (1 + cos 2x)/2 dx = x/2 + sin 2x /4 = (1/2) * (x + sin 2x /2) = F(x)

By second fundamental theorem of calculus, we get

I = F(π/2) – F(0)

= (1/2) * [(π/2 + sin π /2) – (0 + sin 0 /2)

= (1/2) * (π/2 + 0 – 0 - 0)

= π/4

Question 13:

23 [x/(x2 + 1)]dx

Let I = ꭍ23 [x/(x2 + 1)]dx

So, ꭍ [x/(x2 + 1)]dx = (1/2) * ꭍ [2x/(x2 + 1)]dx = (1/2) * log (x2 + 1) = F(x)

By second fundamental theorem of calculus, we get

I = F(3) – F(2)

= (1/2) * [log (32 + 1) - log (22 + 1)]

= (1/2) * [log (9 + 1) – log (4 + 1)]

= (1/2) * [log (10) – log (5)]

= (1/2) * [log (10/5)]

= (1/2) * log 2

Question 14:

01 [(2x + 3)/(5x2 + 1)]dx

Let I = ꭍ01 [(2x + 3)/(5x2 + 1)]dx

So, ꭍ [(2x + 3)/(5x2 + 1)]dx = (1/5) * ꭍ [5(2x + 3)/(5x2 + 1)]dx

= (1/5) * ꭍ [(10x + 15)/(5x2 + 1)]dx

= (1/5) * ꭍ (10x)/(5x2 + 1)]dx + (1/5) * ꭍ [(15)/(5x2 + 1)]dx

= (1/5) * ꭍ (10x)/(5x2 + 1)]dx + 3 * ꭍ dx/(5x2 + 1)

= (1/5) * ꭍ (10x)/(5x2 + 1)]dx + 3 * ꭍ dx/{5(x2 + 1/5)}

= (1/5) * log (5x2 + 1) + (3/5) *{1/(1/√5)} * tan-1 {x/(1/√5)}

= (1/5) * log (5x2 + 1) + (3/√5) * tan-1 (x√5)

= F(x)

By second fundamental theorem of calculus, we get

I = F(1) – F(0)

= [(1/5) * log (5 + 1) + (3/√5) * tan-1 (√5)] – [(1/5) * log (1) + (3/√5) * tan-1 (0)]

= (1/5) * log (6) + (3/√5) * tan-1 (√5)

Question 15:

01 [x * ex2]dx

Let I = ꭍ01 [x * ex2]dx

Put x2 = t

=> 2x dx = dt

=> x dx = dt/2

As x -> 0, t-> 0 and as x -> 1, t-> 1

So, I = ꭍ01 et dt

Now, (1/2) * ꭍ01 et dt = et/2 = F(t)

By second fundamental theorem of calculus, we get

I = F(1) – F(0)

= e1/2 – e0/2

= e/2 – 1/2

= (e - 1)/2

Question 16:

12 [(5x2)/(x2 + 4x + 3)]dx

Let I = ꭍ12 [(5x2)/(x2 + 4x + 3)]dx

Dividing 5x2 by x2 + 4x + 3, we get

I = ꭍ12 [5 – (20x + 15)/(x2 + 4x + 3)]dx

= ꭍ12 5 dx – ꭍ12 [(20x + 15)/(x2 + 4x + 3)]dx

= 5[x]12 – ꭍ12 [(20x + 15)/(x2 + 4x + 3)]dx

= 5[2 - 1] – ꭍ12 [(20x + 15)/(x2 + 4x + 3)]dx

= 5 – ꭍ12 [(20x + 15)/(x2 + 4x + 3)]dx

=> I = 5 – I1     ………………..1

Where I1 = ꭍ12 [(20x + 15)/(x2 + 4x + 3)]dx

Now, consider I1 = ꭍ12 (20x + 15)/(x2 + 4x + 3)]dx

Let 20x + 15 = A * d(x2 + 4x + 3)/dx + B

=> 20x + 15 = A(2x + 4) + B   ………….2

=> 20x + 15 = 2Ax + (4A + B)

Equating the coefficients of x and constant term, we obtain

2A = 20

=> A = 10

4A + B = 15

=> 40 + B = 15

=> B = -25

From equation 2, we get

20x + 15 = 10(2x + 4) – 25

Now, I1 = ꭍ12 (20x + 15)/(x2 + 4x + 3)]dx

= ꭍ12 {10(2x + 4) – 25}/(x2 + 4x + 3)]dx

= 10 * ꭍ12 [(2x + 4)/(x2 + 4x + 3)]dx - 25 * ꭍ12 dx/(x2 + 4x + 3)

Let x2 + 4x + 3 = t

=> (2x + 4)dx = dt

So, I1 = 10 * ꭍ12 dt/t - 25 * ꭍ12 dx/{(x + 2)2 - 12}

=> I1 = 10 * [log t]12 - 25 * (1/2) * [log {(x + 1)/(x + 3)}]12

=> I1 = 10 * [log (x2 + 4x + 3)]12 - 25 * (1/2) * [log {(x + 1)/(x + 3)}]12

=> I1 = 10 * [log (22 + 4 * 2 + 3) - log (12 + 4 * 1 + 3)]

- 25 * (1/2) * [log {(2 + 1)/(2 + 3)} - log {(1 + 1)/(1 + 3)}]

=> I1 = 10 * [log 15 - log 8] - (25/2) * [log (3/5) - log (2/4)]

=> I1 = 10 * [log (5 * 3) - log (4 * 2)] - (25/2) * [log 3 – log 5 - log 2 + log 4]

=> I1 = (10 + 25/2) * log 5 + (-10 - 25/2) * log 4 + (10 - 25/2) * log 3 + (-10 + 25/2) * log 2

=> I1 = (45/2) * log (5/4) - (5/2) * log (3/2)

Substituting the value of I1 in equation 1, we get

=> I = 5 – [(45/2) * log (5/4) - (5/2) * log (3/2)]

=> I = 5 – (5/2) * [9 * log (5/4) - log (3/2)]

Question 17:

0π/4 (2sec2 x + x3 + 2)dx

Let I = ꭍ0π/4 (2sec2 x + x3 + 2)dx

So, ꭍ (2sec2 x + x3 + 2)dx = 2 tan x + x4/4 + 2x = F(x)

By second fundamental theorem of calculus, we get

I = F(π/4) – F(0)

= [2 tan π/4 + (π/4)4/4 + 2 * π/4] – [2 tan 0 + 0 + 0]

= 2 * 1 + π4/45 + π/2

= 2 + π/2 + π4/1024

Question 18:

0π (sin2 x/2 - cos2 x/2)dx

Let I = ꭍ0π (sin2 x/2 - cos2 x/2)dx

= - ꭍ0π (cos2 x/2 - sin2 x/2)dx

= - ꭍ0π cos x dx

Now, ꭍ cos x dx = sin x = F(x)

By second fundamental theorem of calculus, we get

I = F(π) – F(0)

= -[sin π – sin 0]

= 0

Question 19:

02 [(6x + 3)/(x2 + 4)]dx

Let I = ꭍ02 [(6x + 3)/(x2 + 4)]dx

Now, ꭍ [(6x + 3)/(x2 + 4)]dx = 3 * ꭍ [(2x + 1)/(x2 + 4)]dx

= 3 * ꭍ [(2x)/(x2 + 4)]dx + 3 * ꭍ dx/(x2 + 4)

= 3 * log(x2 + 4)] + (3/2) * tan-1 (x/2)

= F(x)

By second fundamental theorem of calculus, we get

I = F(2) – F(0)

= [3 * log(22 + 4) + (3/2) * tan-1 (2/2)] – [3 * log(02 + 4) + (3/2) * tan-1 (0/2)]

= [3 * log(4 + 4) + (3/2) * tan-1 (1)] – [3 * log(0 + 4) + (3/2) * tan-1 (0)]

= [3 * log 8 + (3/2) * (π/4)] – [3 * log 4 + 0]

= 3 * log (8/4) + 3π/8

= 3 * log 2 + 3π/8

Question 20:

01 [x * ex + sin(πx/4)]dx

Let I = ꭍ01 [x * ex + sin(πx/4)]dx

Now, ꭍ [x * ex + sin(πx/4)]dx = x * ꭍ ex dx - ꭍ[d(x)/dx * ꭍ ex dx]dx + [-cos(πx/4)/(π/4)]

= xex - ꭍ ex dx - (4/π) * cos(πx/4)

= xex - ex - (4/π) * cos(πx/4)

= F(x)

By second fundamental theorem of calculus, we get

I = F(1) – F(0)

= [1 * e1 – e1 - (4/π) * cos(π/4)] – [0 * e0 – e0 - (4/π) * cos(0)]

= [e – e - (4/π) * (1/√2)] – [-1 - (4/π)]

= 1 + 4/π - 2√2/π

Choose the correct answer in Exercises 21 and 22.

Question 21:

1√3 dx/(1 + x2) equals

1. π/3 B. 2π/3 C. π/6                                D. π/12

ꭍ dx/(1 + x2) = tan-1 x = F(x)

By second fundamental theorem of calculus, we get

1√3 dx/(1 + x2) = F(√3) – F(1)

= tan-1 √3 - tan-1 1

= π/3 - π/4

= π/12

Hence, the correct answer is option D.

Question 22:

02/3 dx/(4 + 9x2) equals

1. π/6 B. π/12 C. π/24                                D. π/4

ꭍ dx/(4 + 9x2) = ꭍ dx/{22 + (3x)2}

Put 3x = t

=> 3 dx = dt

=> dx = dt/3

So, ꭍ dx/{22 + (3x)2} = (1/3) * ꭍ dt/(22 + t2)

= (1/3) * [(1/2) * tan-1 (t/2)]

= (1/6) * tan-1 (3x/2)

= F(x)

By second fundamental theorem of calculus, we get

02/3 dx/(4 + 9x2) = F(2/3) – F(0)

= (1/6) * tan-1 {(3/2) * (2/3)} - (1/6) * tan-1 0

= (1/6) * tan-1 1 – 0

= (1/6) * (π/4)

= π/24

Hence, the correct answer is option C.

Exercise 7.10

Evaluate the integrals in Exercises 1 to 8 using substitution.

Question 1:

ʃ01 [x/(x2 + 1)] dx

Let x2 + 1 = t

=> 2x dx = dt

=> x dx = dt/2

When x = 0, t = 1 and when x = 1, t = 2

So, ʃ01 [x/(x2 + 1)] dx = (1/2)ʃ12 dt/t

= (1/2)[log t]12

= (1/2)[log 2 – log 1]

= (log 2)/2

Question 2:

ʃ0π/2 [√(sin ф) * cos5 ф] dф

Let I = ʃ0π/2 [√(sin ф) * cos5 ф] dф

= ʃ0π/2 [√(sin ф) * cos4 ф * cos ф] dф

Let sin ф = t

=> cos ф dф = dt

When ф = 0, t = 0 and when ф = π/2, t = 1

So, I = ʃ01 [√t * (1 – t2)2] dt

= ʃ01 [√t * (1 + t4 - 2t2)] dt

= ʃ01 [t1/2 + t9/2 - 2t5/2] dt

= [t3/2/(3/2) + t11/2/(11/2) - 2t7/2/(7/2)]01

= 1/(3/2) + 1/(11/2) - 2/(7/2)

= 2/3 + 2/11 – 4/7

= (154 + 42 - 132)/231

= 64/231

Question 3:

ʃ01 [sin-1{2x/(1 + x2)}] dx

Let I = ʃ01 [sin-1{2x/(1 + x2)}] dx

Again, let x = tan θ

=> dx = sec2 θ dθ

When x = 0, θ = 0 and when x = 1, θ = π/4

So, I = ʃ0π/4 [sin-1{2tan θ /(1 + tan2 θ)} * sec2 θ] dθ

= ʃ0π/4 [sin-1{sin 2θ} * sec2 θ] dθ

= ʃ0π/4 [2θ * sec2 θ] dθ

= 2 * ʃ0π/4 [θ * sec2 θ] dθ

Taking θ as first function and sec2 θ as second function and integrating by parts, we get

So, I = 2[θ * ʃ sec2 θ dθ - ʃ {(dθ/dθ) * ʃ sec2 θ dθ} dθ]0π/4

=> I = 2[θ * tan θ - ʃ tan θ dθ]0π/4

=> I = 2[θ * tan θ + log|cos θ|]0π/4

=> I = 2[π/4 * tan θ π/4 + log|cos π/4| - log|cos 0|]

=> I = 2[π/4 + log(1/√2) – log 1]

=> I = 2[π/4 + (1/2) log(1/2)]

=> I = π/2 – log 2

Question 4:

ʃ02 [x√(x + 2)] dx              (put x + 2 = t2)

Let x + 2 = t2

=> dx = 2t dt

When x = 0, t = √2 and when x = 2, t = 2

So, ʃ02 [x√(x + 2)] dx = ʃ22 [(t2 - 2) * √(t2) * 2t] dt

= 2 * ʃ22 [(t2 - 2) * t2] dt

= 2 * ʃ22 [t4 - 2t2] dt

= 2[t5/5 – 2t3/3]22

= 2[32/5 – 16/3 – 4√2/4 + 4√2/3]

= 2[(96 – 80 – 12√2 + 20√2)/15]

= 2[(16 + 8√2)/15]

= 16[(2 + √2)/15]

= [16√2(√2 + 1)]/15

Question 5:

ʃ0π/2 [sin x /(1 + cos2 x)] dx

Let cos x = t

=> -sin x dx = dt

When x = 0, t = 1 and when x = π/2, t = 0

So, ʃ0π/2 [sin x /(1 + cos2 x)] dx = - ʃ10 [dt/(1 + t2)] dt

= -[tan-1 t]10

= -[tan-1 0 - tan-1 1]

= -(-π/4)

= π/4

Question 6:

ʃ02 [dx/(x + 4 - x2)]

ʃ02 [dx/(x + 4 - x2)] = ʃ02 [-dx/(x2 – x – 4)]

= ʃ02 [-dx/(x2 – x + 1/4 – 1/4 - 4)]

= ʃ02 [-dx/{(x – 1/2)2 – 17/4}]

= ʃ02 [-dx/{(x – 1/2)2 – (√17/2)2}]

= ʃ02 [dx/{(√17/2)2 - (x – 1/2)2}]

Let x – 1/2 = t

=> dx = dt

When x = 0, t = -1/2 and when x = 2, t = 3/2

So, ʃ02 [dx/{(√17/2)2 - (x – 1/2)2}] = ʃ-1/23/2 [dt/{(√17/2)2 - t2}]

= [1/{2 * √17/2} * log{(√17/2 + t)/(√17/2 - t)}]-1/23/2

= (1√17) * [log{(√17/2 + 3/2)/(√17/2 – 3/2)}

- log{(√17/2 – 1/2)/(√17/2 + 1/2)}]

= (1√17) * [log{(√17 + 3)/(√17 – 3)} - log{(√17 – 1)/(√17 + 1)}]

= (1√17) * [log{(√17 + 3)/(√17 – 3) * (√17 + 1)/(√17 - 1)}]

= (1√17) * [log{(17 + 3 + 4√17)/(17 + 3 - 4√17)}]

= (1√17) * [log{(20 + 4√17)/(20 - 4√17)}]

= (1√17) * [log{(5 + √17)/(5 - √17)}]

= (1√17) * [log{(5 + √17)* (5 + √17)/(5 - √17)* (5 + √17)}]

= (1√17) * [log{(25 + 17 + 10√17)/(25 - 17)}]

= (1√17) * [log{(42 + 10√17)/8}]

= (1√17) * [log{(21 + 5√17)/4}]

Question 7:

ʃ-11 [dx/(x2 + 2x + 5)]

ʃ-11 [dx/(x2 + 2x + 5)] = ʃ-11 [dx/{(x2 + 2x + 1) + 4}] = ʃ-11 [dx/{(x + 1)2 + 22}]

Let x + 1 = t

=> dx = dt

When x = -1, t = 0 and when x = 1, t = 2

So, ʃ-11 [dx/{(x + 1)2 + 22}] = ʃ02 [dt/(t2 + 22)]

= (1/2)[tan-1 (t/2)]02

= (1/2)[tan-1 (2/2) - tan-1 (0/2)]

= (1/2)[tan-1 1]

= (1/2) * (π/4)

= π/8

Question 8:

ʃ12 [(1/x – 1/2x2) * e2x]dx

Let 2x = t

=> 2 dx = dt

=> dx = dt/2

When x = 1, t = 2 and when x = 2, t = 4

So, ʃ12 [(1/x – 1/2x2) * e2x]dx = (1/2)ʃ24 [(2/t – 2/t2) * et]dt

= ʃ24 [(1/t – 1/t2) * et]dt

Let 1/t = f(t)

=> f’(t) = -1/t2

So, ʃ24 [(1/t – 1/t2) * et]dt = ʃ24 [et{f(t) + f’(t)]dt

= [et * f(t)]24

= [et * 1/t]24

= [et/t]24

= [e4/4 - e2/2]

= e2(e2 – 2)/4

Choose the correct answer in Exercises 9 and 10.

Question 9:

The value of the integral ʃ1/31 [(x – x3)1/3/x4]dx

(A) 6                                        (B) 0                                          (C) 3                                    (D) 4

Let I = ʃ1/31 [(x – x3)1/3/x4]dx

Also, let x = sin θ

=> dx = cos θ dθ

When x = 1/3, θ = sin-1(1/3) and when x = 1, θ = π/2

So, I = ʃsin-1(1/3)π/2 [{(sin θ – sin3 θ)1/3/sin4 θ} * cos θ]dθ

= ʃsin-1(1/3)π/2 [{(sin θ)1/3(1  – sin2 θ)1/3/sin4 θ} * cos θ]dθ

= ʃsin-1(1/3)π/2 [{(sin θ)1/3(cos2 θ)1/3/sin4 θ} * cos θ]dθ

= ʃsin-1(1/3)π/2 [{(sin θ)1/3(cos θ)2/3/sin4 θ} * cos θ]dθ

= ʃsin-1(1/3)π/2 [{(sin θ)1/3(cos θ)2/3/(sin2 θ * sin2 θ)} * cos θ]dθ

= ʃsin-1(1/3)π/2 [{(cos θ)5/3/(sin θ)5/3} * cosec2 θ]dθ

= ʃsin-1(1/3)π/2 [(cot θ)5/3 * cosec2 θ]dθ

Let cot θ = t

=> -cosec2 θ dθ = dt

=> cosec2 θ dθ = -dt

When θ = sin-1(1/3), t = 2√2 and when θ = π/2, t = 0

So, I = -ʃ220 t5/3 dt

=> I = -[t8/3/(8/3)]220

=> I = -[3t8/3/8]220

=> I = -[0 – 3(2√2)8/3/8]

=> I = 3(√8)8/3/8

=> I = 3(8)4/3/8

=> I = 3(23)4/3/8

=> I = 3(2)4/8

=> I = (3 * 16)/8

=> I = 3 * 2

=> I = 6

Hence, the correct answer is option A.

Question 10:

If f(x) = ʃ0x t * sin t dt, then f’(x) is

1. cos x + x sin x B. x sin x C. x cos x                       D. sin x + x cos x

Given, f(x) = ʃ0x t * sin t dt

Integrating by parts, we obtain

f(x) = t ʃ0x sin t dt - ʃ0x [(dt/dt) * ʃ sin t dt]dt

= [-t * cos t]0x - ʃ0x (-cos t dt)

= [-t * cos t]0x + ʃ0x cos t dt

= [-t * cos t + sin t]0x

= -x * cos x + sin x

Now, f’(x) = -[x(-sin x) + cos x] + cos x

= x sin x - cos x + cos x

= x sin x

Hence, the correct answer is option B.

Exercise 7.11

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Question 1:

ʃ0π/2 cos2 x dx

Let I = ʃ0π/2 cos2 x dx    …………..1

=> I = ʃ0π/2 cos2 (π/2 - x) dx

=> I = ʃ0π/2 sin2 x dx      …………..2                       [Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

Adding equation 1and 2, we get

2I = I = ʃ0π/2 (sin2 x + cos2 x) dx

=> 2I = I = ʃ0π/2 1 dx

=> 2I = I = [x]0π/2

=> 2I = π/2

=> I = π/4

So, ʃ0π/2 cos2 x dx = π/4

Question 2:

ʃ0π/2 [√sin x/(√sin x + √cos x)] dx

Let I = ʃ0π/2 [√sin x/(√sin x + √cos x)] dx   …………….1

=> I = ʃ0π/2 [√sin (π/2 - x)/(√sin (π/2 - x) + √cos (π/2 - x))] dx            [Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

=> I = ʃ0π/2 [√cos x/(√cos x + √sin x)] dx   …………….2

Adding equation 1and 2, we get

2I = ʃ0π/2 [(√sin x + √cos x)/(√cos x + √sin x)] dx

=> 2I = I = ʃ0π/2 1 dx

=> 2I = I = [x]0π/2

=> 2I = π/2

=> I = π/4

So, ʃ0π/2 [√sin x/(√sin x + √cos x)] dx = π/4

Question 3:

ʃ0π/2 [sin3/2 x/(sin3/2 x + cos3/2 x)] dx

Let I = ʃ0π/2 [sin3/2 x/(sin3/2 x + cos3/2 x)] dx   …………….1

=> I = ʃ0π/20π/2 [sin3/2 (π/2 - x)/{sin3/2 (π/2 - x) + cos3/2 (π/2 - x))] dx

[Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

=> I = ʃ0π/2 [cos3/2 x/([cos3/2 x + sin3/2 x)] dx   …………….2

Adding equation 1and 2, we get

2I = ʃ0π/2 [(sin3/2 x + cos3/2 x)/( cos3/2 x + sin3/2 x)] dx

=> 2I = I = ʃ0π/2 1 dx

=> 2I = I = [x]0π/2

=> 2I = π/2

=> I = π/4

So, ʃ0π/2 [sin3/2 x/(sin3/2 x + cos3/2 x)] dx = π/4

Question 4:

ʃ0π/2 [cos5 x/(sin5 x + cos5 x)] dx

Let I = ʃ0π/2 [cos5 x/(sin5 x + cos5 x)] dx   …………….1

=> I = ʃ0π/20π/2 [cos5 (π/2 - x)/{sin5 (π/2 - x) + cos5 (π/2 - x))] dx

[Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

=> I = ʃ0π/2 [sin5 x/([cos5 x + sin5 x)] dx   …………….2

Adding equation 1and 2, we get

2I = ʃ0π/2 [(sin5 x + cos5 x)/(cos5 x + sin5 x)] dx

=> 2I = I = ʃ0π/2 1 dx

=> 2I = I = [x]0π/2

=> 2I = π/2

=> I = π/4

So, ʃ0π/2 [cos5 x/(sin5 x + cos5 x)] dx = π/4

Question 5:

ʃ-55 |x + 2| dx

Let I = ʃ-55 |x + 2| dx

It can be seen that (x + 2) ≤ 0 on [−5, −2] and (x + 2) ≥ 0 on [−2, 5].

So, I = ʃ-5-2 {-(x + 2)} dx + ʃ-25 (x + 2) dx

=> I = -ʃ-5-2 (x + 2) dx + ʃ-25 (x + 2) dx

=> I = -[x2/2 + 2x]-5-2 + [x2/2 + 2x]-25

=> I = -[(-2)2/2 + 2 * (-2) - (-5)2/2 - 2 * (-5)] + [52/2 + 2 * 5 - (-2)2/2 - 2 * (-2)]

=> I = -[2 - 4 - 25/2 + 10] + [25/2 + 10 - 2 + 4]

=> I = -2 + 4 + 25/2 - 10] + 25/2 + 10 - 2 + 4

=> I = 29

So, ʃ-55 |x + 2| dx = 29

Question 6:

ʃ28 |x - 5| dx

Let I = ʃ28 |x - 5| dx

It can be seen that (x - 5) ≤ 0 on [2, 5] and (x - 5) ≥ 0 on [5, 8].

So, I = ʃ25 {-(x - 5)} dx + ʃ58 (x - 5) dx

=> I = - ʃ25 (x - 5) dx + ʃ58 (x - 5) dx

=> I = -[x2/2 - 5x]25 + [x2/2 - 5x]58

=> I = -[52/2 - 5 * 5 - 22/2 + 5 * 2] + [82/2 - 5 * 8 - 52/2 + 5 * 5]

=> I = -[25/2 - 25 - 2 + 10] + [32 - 40 – 25/2 + 25]

=> I = 9

So, ʃ28 |x - 5| dx = 9

Question 7:

ʃ01 {x(1 - x)n} dx

Let I = ʃ01 {x(1 - x)n} dx

=> I = ʃ01 [(1 - x){1 - (1 - x)}n] dx                    [Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

=> I = ʃ01 [(1 - x)* xn] dx

=> I = ʃ01 [xn - xn+1] dx

=> I = [xn+1/(n + 1)- xn+2/(n + 2)]01

=> I = [1/(n + 1)- 1/(n + 2)]

=> I = {(n + 2) – (n + 1)}/{(n + 1)*(n + 2)}

=> I = 1/{(n + 1)*(n + 2)}

So, ʃ01 {x(1 - x)n} dx = 1/{(n + 1)*(n + 2)}

Question 8:

ʃ0π/4 log(1 + tan x) dx

Let I = ʃ0π/4 log(1 + tan x) dx     …………..1

=> I = ʃ0π/4 log{1 + tan (π/4 - x)} dx                                 [Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

=> I = ʃ0π/4 log{1 + (tan π/4 – tan x)/(1 + tan π/4 * tan x)} dx

=> I = ʃ0π/4 log{1 + (1 – tan x)/(1 + tan x)} dx

=> I = ʃ0π/4 log{(1 + tan x + 1 – tan x)/(1 + tan x)} dx

=> I = ʃ0π/4 log{2/(1 + tan x)} dx

=> I = ʃ0π/4 log 2 dx - ʃ0π/4 (1 + tan x)} dx

=> I = log 2 * ʃ0π/4 dx – I                              [From equation 1]

=> 2I = log 2 * [x]0π/4

=> 2I = log 2 * [π/4]

=> I = (π/8) * log 2

So, ʃ0π/4 log(1 + tan x) dx = (π/8) * log 2

Question 9:

ʃ02 x√(2 - x) dx

Let I = ʃ02 x√(2 - x) dx

=> I = ʃ02 [(2 - x)√{2 - (2 - x)}] dx

=> I = ʃ02 [(2 - x)√x] dx                             [Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

=> I = ʃ02 [2√x - x√x] dx

=> I = ʃ02 [2x1/2 – x3/2] dx

=> I = [2x3/2/(3/2) – x5/2/(5/2)]02

=> I = [2 * 23/2/(3/2) – 25/2/(5/2)]

=> I = (4/3) * 23/2 – (2/5) * 25/2

=> I = (4 * 2√2)/3 – (2 * 4√2)/5

=> I = 8√2/3 – 8√2/5

=> I = (40√2 – 24√2)/15

=> I = 16√2/15

So, ʃ02 x√(2 - x) dx = 16√2/15

Question 10:

ʃ0π/2 (2log sin x – log sin 2x)dx

Let I = ʃ0π/2 (2log sin x – log sin 2x)dx

=> I = ʃ0π/2 {2log sin x – log (2 * sin x * cos x}dx

=> I = ʃ0π/2 {2log sin x – log sin x – log cos x – log 2}dx

=> I = ʃ0π/2 {log sin x – log cos x – log 2}dx     ………….1

=> I = ʃ0π/2 {log sin (π/2 - x) – log cos (π/2 - x) – log 2}dx              [Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

=> I = ʃ0π/2 {log cos x – log sin x – log 2}dx     ………….2

Adding equations 1 and 2, we get

=> 2I = ʃ0π/2 (-log 2 – log 2)dx

=> 2I = ʃ0π/2 (-2log 2)dx

=> 2I = (-2log 2) * ʃ0π/2 dx

=> 2I = (-2log 2) * [x]0π/2

=> 2I = (-2log 2) * (π/2)

=> I = log (1/2) * (π/2)

=> I = (π/2) * log (1/2)

So, ʃ0π/2 (2log sin x – log sin 2x)dx = (π/2) * log (1/2)

Question 11:

ʃ-π/2π/2 sin2 x dx

Let I = ʃ-π/2π/2 sin2 x dx

Since sin2 (−x) = {sin(−x)}2 = (−sin x)2 = sin2 x, therefore, sin2 x is an even function.

It is known that if f(x) is an even function, then ʃ-aa f(x) dx = 2 * ʃ0a f(x) dx

So, I = 2 * ʃ0π/2 sin2 x dx

=> I = 2 * ʃ0π/2 {(1 – cos 2x)/2} dx

=> I = ʃ0π/2 (1 – cos 2x) dx

=> I = [x – sin 2x /2]0π/2

=> I = [π/2 – sin (2 * π/2) /2]

=> I = [π/2 – (sin π)/2]

=> I = π/2

So, ʃ-π/2π/2 sin2 x dx = π/2

Question 12:

ʃ0π [x/(1 + sin x)] dx

Let I = ʃ0π [x/(1 + sin x)] dx            …………..1

=> I = ʃ0π [(π - x)/{1 + sin (π - x)}] dx                             [Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

=> I = ʃ0π [(π - x)/(1 + sin x)] dx     ………….2

Adding equatns 1 and 2, we get

=> 2I = ʃ0π [π/(1 + sin x)] dx

=> 2I = ʃ0π [{π(1 - sin x)}/{(1 + sin x)(1 - sin x)}] dx

=> 2I = ʃ0π [{π(1 - sin x)}/(1 – sin2 x)] dx

=> 2I = ʃ0π [{π(1 - sin x)}/cos2 x] dx

=> 2I = π * ʃ0π [sec2 x – tan x * sec x] dx

=> 2I = π * [tan x – sec x]0π

=> 2I = π * [(tan π – sec π) – (tan 0 – sec 0)]

=> 2I = π * [1 + 1]

=> 2I = π * 2

=> I = π

So, ʃ0π [x/(1 + sin x)] dx = π

Question 13:

ʃ-π/2π/2 sin7 x dx

Let I = ʃ-π/2π/2 sin7 x dx

Since sin7 (−x) = {sin(−x)}7 = (−sin x)7 = -sin7 x, therefore, sin7 x is an odd function.

It is known that if f(x) is an odd function, then ʃ-aa f(x) dx = 0

So, I = ʃ-π/2π/2 sin7 x dx = 0

Question 14:

ʃ0 cos5 x dx

Let I = ʃ0 cos5 x dx

Now, cos5 (2π - x) = cos5 x

It is known that

ʃ02a f(x) dx =   2 * ʃ0a f(x) dx, if f(2a - x) = f(x)

0, if f(2a - x) = -f(x)

So, I = 2 * ʃ0π cos5 x dx

=> I = 2 * 0                              [Since cos5 (π - x) = -cos5 x]

=> I = 0

So, ʃ0 cos5 x dx = 0

Question 15:

ʃ0π/2 [(sin x – cos x)/(1 + sin x * cos x)]dx

Let I = ʃ0π/2 [(sin x – cos x)/(1 + sin x * cos x)]dx     ………….1

=> I = ʃ0π/2 [{sin (π/2 - x) – cos (π/2 - x)}/{1 + sin (π/2 - x) * cos (π/2 - x)}]dx

[Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

=> I = ʃ0π/2 [(cos x – sin x)/(1 + sin x * cos x)]dx      ………….2

Adding equations 1 and 2, we get

=> 2I = ʃ0π/2 [(sin x – cos x + cos x – sin x)/(1 + sin x * cos x)]dx

=> 2I = 0

=> I = 0

So, ʃ0π/2 [(sin x – cos x)/(1 + sin x * cos x)]dx = 0

Question 16:

ʃ0π log(1 + cos x)dx

Let I = ʃ0π log(1 + cos x)dx     …………1

=> I = ʃ0π log{1 + cos (π - x)}dx                         [Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

=> I = ʃ0π log(1 - cos x)dx         ………….2

Adding equations 1 and 2, we get

=> 2I = ʃ0π {log(1 + cos x) + log(1 - cos x)}dx

=> 2I = ʃ0π [log{(1 + cos x) * log(1 - cos x)}]dx

=> 2I = ʃ0π log(1 – cos2 x) dx

=> 2I = ʃ0π log sin2 x dx

=> 2I = ʃ0π 2 * log sin x dx

=> I = ʃ0π log sin x dx        ………….3

Now, sin(π - x) = sin x

=> I = 2 * ʃ0π/2 log sin x dx           …………4

=> I = 2 * ʃ0π/2 log sin (π/2 - x) dx

=> I = 2 * ʃ0π/2 log cos x dx          …………5

Adding equations 4 and 5, we get

=> 2I = 2 * ʃ0π/2 (log sin x + log cos x)dx

=> I = ʃ0π/2 (log sin x + log cos x + log 2 – log 2)dx

=> I = ʃ0π/2 {log (2 * sin x * log cos x) – log 2}dx

=> I = ʃ0π/2 {log sin 2x – log 2}dx

=> I = ʃ0π/2 {log sin 2x – log 2}dx

=> I = ʃ0π/2 log sin 2x dx – ʃ0π/2 log 2 dx    …………6

Let 2x = t

=> 2 dx = dt

=> dx = dt/2

When x = 0, t = 0 and when x = π/2, t = π

From equation 6, we get

=> I = (1/2) * ʃ0π log sin t dt – ʃ0π/2 log 2 dx

=> I = (1/2) * ʃ0π log sin x dx – ʃ0π/2 log 2 dx

=> I = I/2 – log 2 * ʃ0π/2 dx                                       [From equation 3]

=> I = I/2 – log 2 * [x]0π/2

=> I - I/2 = – (π/2) * log 2

=> I/2 = – (π/2) * log 2

=> I = -π * log 2

So, ʃ0π log(1 + cos x)dx = -π * log 2

Question 17:

ʃ0a [√x/{√x + √(a - x)}]dx

Let I = ʃ0a [√x/{√x + √(a - x)}]dx     …………1

=> I = ʃ0a [√(a - x)/{√(a - x) + √(a – (a - x))}]dx                   [Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

=> I = ʃ0a [√(a - x)/{√(a - x) + √x}]dx   ………2

Adding equations 1 and 2, we get

=> 2I = ʃ0a [{√x + √(a - x)}/{√(a - x) + √x}]dx

=> 2I = ʃ0a dx

=> 2I = [x]0a

=> 2I = a

=> I = a/2

So, ʃ0a [√x/{√x + √(a - x)}]dx = a/2

Question 18:

ʃ04 |x - 1|dx

Let I = ʃ04 |x - 1|dx

It can be seen that, (x − 1) ≤ 0 when 0 ≤ x ≤ 1 and (x − 1) ≥ 0 when 1 ≤ x ≤ 4

So, I = ʃ01 |x - 1|dx + ʃ14 |x - 1|dx                               [Since ʃab f(x) dx = ʃac f(x) dx + ʃcb f(x) dx]

=> I = ʃ01 {-(x - 1)} dx + ʃ14 (x – 1) dx

=> I = -ʃ01 (x - 1) dx + ʃ14 (x – 1) dx

=> I = - [x2/2 – x]01 + [x2/2 – x]14

=> I = - [1/2 – 1] + [(42/2 – 4) – (1/2 - 1)]

=> I = - [1/2 – 1] + [8 – 4 – 1/2 + 1]

=> I = 1 – 1/2 + 8 – 4 – 1/2 + 1

=> I = 5

So, ʃ04 |x - 1|dx = 5

Question 19:

Show that ʃ0a f(x) * g(x) dx = 2 * ʃ0a f(x) dx, if f and g are defined as f(x) = f(a – x) and                  g(x) + g(a – x) = 4

Let I = ʃ0a f(x) * g(x) dx     …………..1

=> I = ʃ0a f(a - x) * g(a - x) dx                         [Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

=> I = ʃ0a f(x) * g(a - x) dx    ………..2

Adding equations 1 and 2, we get

=> 2I = ʃ0a [f(x) * g(x) + f(x) * g(a - x)] dx

=> 2I = ʃ0a f(x)[g(x) + g(a - x)] dx

=> 2I = ʃ0a [f(x) * 4] dx                           [Given, g(x) + g(a – x) = 4]

=> I = 2 * ʃ0a f(x) dx

Hence, proved.

Choose the correct answer in Exercises 20 and 21.

Question 20:

The value of ʃ-π/2π/2 (x3 + x * cos x + tan5 x + 1) dx is

1. 0 B. 2 C. π                                     D. 1

Let I = ʃ-π/2π/2 (x3 + x * cos x + tan5 x + 1) dx

=> I = ʃ-π/2π/2 x3 dx + ʃ-π/2π/2 x * cos x dx + ʃ-π/2π/2 tan5 x dx + ʃ-π/2π/2 1 dx

It is known that if f(x) is an even function, then ʃ-aa f(x) dx = 2 * ʃ0a f(x) dx

And if f ( x ) is an odd function, then ʃ-aa f(x) dx = 0

So, I = 0 + 0 + 0 + 2 *  ʃ0π/2 1 dx

=> I = 2 * [x]0π/2

=> I = 2 * (π/2)

=> I = π

Hence, the correct anser is option C.

Question 21:

The value of ʃ0π/2 log{(4 + 3 * sin x)/(4 + 3 * cos x)} dx is

1. 2 B. 3/4 C. 0                                     D. -2

Let I = ʃ0π/2 log{(4 + 3 * sin x)/( 4 + 3 * cos x)} dx    …………..1

=> I = ʃ0π/2 log[{4 + 3 * sin (π/2 - x)}/{4 + 3 * cos (π/2 - x)}] dx        [Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

I = ʃ0π/2 log{(4 + 3 * cos x)/( 4 + 3 * sin x)} dx      ………….2

Adding equations 1 and 2, we get

=> 2I = ʃ0π/2 [log{(4 + 3 * sin x)/( 4 + 3 * cos x)} + log{(4 + 3 * cos x)/( 4 + 3 * sin x)}] dx

=> 2I = ʃ0π/2 [log{(4 + 3 * sin x)/( 4 + 3 * cos x) * (4 + 3 * cos x)/( 4 + 3 * sin x)}] dx

=> 2I = ʃ0π/2 log 1 dx

=> 2I = ʃ0π/2 0 dx

=> 2I = 0

=> I = 0

Hence, the correct answer is option C.

Miscellaneous Exercise of chapter 7

Integrate the functions in Exercises 1 to 24.

Question 1:

1/(x – x3)

1/(x – x3) = 1/{x(1 – x2)} = 1/{x(1 - x)(1 + x)}

Let 1/{x(1 - x)(1 + x)} = A/x + B/(1 - x) + C/(1 + x)    …………..1

=> 1 = A(1 - x)(1 + x) + Bx(1 + x) + Cx(1 - x)

=> 1 = A(1 – x2) + Bx(1 + x) + Cx(1 - x)

Equating the coefficients of x2, x, and constant term, we get

-A + B - C = 0

B + C = 0

A = 1

On solving these equations, we obtain

A = 1, B = 1/2 and C = -1/2

From equation 1, we get

1/{x(1 - x)(1 + x)} = 1/x + 1/2(1 - x) - 1/2(1 + x)

Now, ꭍ [1/{x(1 - x)(1 + x)}] dx = ꭍ dx/x + (1/2) * ꭍ dx/(1 - x) – (1/2) * ꭍ dx/(1 + x)

= log|x| - (1/2) * log|1 - x| – (1/2) * log|1 + x| + C

= log|x| - log|(1 – x)1/2| – log|(1 + x)1/2| + C

= log|x| - log|1/{(1 – x)1/2/(1 + x)1/2| + C

= log|x| - [log|1/{(1 – x2)1/2|] + C

= log|x/{(1 – x2)1/2| + C

= log|{x2/(1 – x2)}1/2| + C

= (1/2) * log|x2/(1 – x2)| + C

Question 2:

1/{√(x + a) + √(x + b)}

1/{√(x + a) + √(x + b)} = [1/{√(x + a) + √(x + b)}] * [{√(x + a) - √(x + b)}/{√(x + a) - √(x + b)}]

= {√(x + a) - √(x + b)}/{(x + a) - (x + b)}

= {√(x + a) - √(x + b)}/(a - b)

Now, ꭍ dx/{√(x + a) + √(x + b)} = 1/(a - b) * ꭍ {√(x + a) - √(x + b)} dx

= 1/(a - b) * ꭍ {(x + a)1/2 - (x + b)1/2} dx

= 1/(a - b) * {(x + a)3/2/(3/2) - (x + b)3/2/(3/2)} + C

= 2/3(a - b) * [(x + a)3/2 - (x + b)3/2] + C

Question 3:

1/{x√(ax – x2)}                     [Hint Put x = a/t]

Let x = a/t

=> dx = -a/t2

Now, ꭍ dx/{x√(ax – x2)} = ꭍ dt/[(a/t) * √{a * (a/t) – (a/t)2}] * (-a/t2)

= -ꭍ dt/[1/(at) * √(1/t – 1/t2)]

= (-1/a) * ꭍ dt/[√(t2/t – t2/t2)

= (-1/a) * ꭍ dt/[√(t – 1)

= (-1/a) * 2√(t – 1) + C

= (-1/a) * 2[√(a/x – 1)] + C

= (-2/a) * [√(a – x)/√x] + C

= (-2/a) * √{(a – x)/x} + C

Question 4:

1/{x2(x4 + 1)3/4}

Multiplying and divide by x-3, we get

1/{x2(x4 + 1)3/4} = x-3/{x-3 * x2(x4 + 1)3/4}

= {x-3 * (x4 + 1)-3/4}/(x-3 * x2)

= (x4 + 1)/{x5 * (x4)-3/4}

= (1/x5) * {(x4 + 1)/x4}-3/4

Let 1/x4 = t

=> (-4/x5) dx= dt

=> (1/x5) dx= dt/(-4)

So, ꭍ dx/{x2(x4 + 1)3/4} = (-1/4) * ꭍ (1 + t)-3/4 dt

= (-1/4) * [(1 + t)1/4/(1/4)] + C

= -(1 + t)1/4 + C

= -(1 + 1/x4)1/4 + C

Question 5:

1/(x1/2 + x1/3)                                           [Hint: 1/(x1/2 + x1/3) = 1/{x1/3(1 + x1/6)}, Put x = t6]

1/(x1/2 + x1/3) = 1/{x1/3(1 + x1/6)}

Let Put x = t6

=> dx = 6t5 dt

Now, ꭍ dx/(x1/2 + x1/3) = ꭍ dx/{x1/3(1 + x1/6)}

= ꭍ [6t5/{t2(1 + t)}] dt

= 6 * ꭍ [t3/(1 + t)] dt

On dividing, we get

ꭍ dx/(x1/2 + x1/3) = 6 * ꭍ [(t2 - t + 1) – 1/(1 + t)] dt

= 6 * [t3/3 – t2/2 + t – log|1 + t|] + C

= 2x1/2 – 3x1/3 + 6x1/6 – 6 * log|1 + x1/6|] + C

= 2√x – 3x1/3 + 6x1/6 – 6 * log|1 + x1/6|] + C

Question 6:

5x/{(x + 1)(x2 + 9)}

Let 5x/{(x + 1)(x2 + 9)} = A/(x + 1) + (Bx + C)/(x2 + 9)      …………..1

=> 5x = A(x2 + 9) + (Bx + C)/(x + 1)

=> 5x = Ax2 + 9a + Bx2 + Bx + Cx + C

Equating the coefficients of x2, x, and constant term, we get

A + B = 0

B + C = 5

9A + C = 0

On solving these equations, we obtain

A = -1/2, B = 1/2 and C = 9/2

From equation 1, we get

5x/{(x + 1)(x2 + 9)} = -1/2(x + 1) + (x/2 + 9/2)/(x2 + 9)

= -1/2(x + 1) + (x + 9)/2(x2 + 9)

Now, ꭍ [5x/{(x + 1)(x2 + 9)}] dx

= ꭍ [-1/2(x + 1) + (x + 9)/2(x2 + 9)] dx

= (-1/2) * log|x + 1| + (1/2) * ꭍ [x/(x2 + 9)] dx + (9/2) * ꭍ dx/(x2 + 9)

= (-1/2) * log|x + 1| + (1/4) * ꭍ [2x/(x2 + 9)] dx + (9/2) * ꭍ dx/(x2 + 9)

= (-1/2) * log|x + 1| + (1/4) * log|x2 + 9| + (9/2) * (1/3) * tan-1 (x/3) + C

= (-1/2) * log|x + 1| + (1/4) * log|x2 + 9| + (3/2) * tan-1 (x/3) + C

Question 7:

sin x /sin(x - a)

Let x – a = t

=> dx = dt

So, ꭍ [sin x /sin(x - a)]dx = ꭍ [sin (t + a) /sin t] dt

= ꭍ [(sin t * cos a + cos t * sin a) /sin t] dt

= ꭍ [cos a + cot t * sin a] dt

= t * cos a + sin a * log|sin t| + C1

= (x - a) * cos a + sin a * log|sin (x - a)| + C1

= x * cos a + sin a * log|sin (x - a)| - a * cos a + C1

= x * cos a + sin a * log|sin (x - a)|+ C

Question 8:

(e5 log x – e4 log x)/(e3 log x – e2 log x)

(e5 log x – e4 log x)/(e3 log x – e2 log x) = [e4 log x(elog x – 1)/e2 log x (elog x – 1)]

= e4 log x/e2 log x

= e2 log x

= elog x2

= x2

So, ꭍ [(e5 log x – e4 log x)/(e3 log x – e2 log x)] dx = ꭍ x2 dx = x3/3 + C

Question 9:

cos x /√(4 – sin2 x)

Let sin x = t

=> cos x dx = dt

Now, ꭍ [cos x /√(4 – sin2 x)] dx = ꭍ dt/√(22 – t2)

= sin-1 (t/2) + C

= sin-1 (sin x /2) + C

Question 10:

(sin8 x – cos8 x)/(1 – 2 * sin2 x * cos2 x)

Given, (sin8 x – cos8 x)/(1 – 2 * sin2 x * cos2 x)

= [(sin4 x – cos4 x)(sin4 x + cos4 x)]/( sin2 x + cos2 x – sin2 x * cos2 x - sin2 x * cos2 x)

= [(sin4 x + cos4 x)(sin2 x + cos2 x)(sin2 x – cos2 x)]/(sin2 x + cos2 x – sin2 x * cos2 x - sin2 x * cos2 x)

= [(sin4 x + cos4 x)(sin2 x – cos2 x)]/[(sin2 x(1 - cos2 x) + cos2 x(1 - sin2 x)]

= [(sin4 x + cos4 x)(sin2 x – cos2 x)]/(sin2 x * sin2 x + cos2 x * cos2 x]

= [(sin4 x + cos4 x)(sin2 x – cos2 x)]/(sin4 x + cos4 x)

= - (cos2 x – sin2 x)

= -cos 2x

So, ꭍ [(sin8 x – cos8 x)/(1 – 2 * sin2 x * cos2 x)] dx = -ꭍ cos 2x dx = -(sin 2x)/2 + C

Question 11:

1/{cos(x + a) * cos(x + b)}

Multiplying and dividing by sin(a - b), we get

1/{cos(x + a) * cos(x + b)}

= {1/sin(a - b)} * [sin(a - b)/{cos(x + a) * cos(x + b)}]

= {1/sin(a - b)} * [sin{(x + a) – (x + b)}/{cos(x + a) * cos(x + b)}]

= {1/sin(a - b)} * [{sin(x + a) * cos(x + b) - cos(x + a) * sin(x + b)}/{cos(x + a) * cos(x + b)}]

= {1/sin(a - b)} * [sin(x + a)/cos(x + a) - sin(x + b)/cos(x + b)]

= {1/sin(a - b)} * [tan{(x + a) - tan(x + b)]

So, ꭍ dx/{cos(x + a) * cos(x + b)} = {1/sin(a - b)} * ꭍ [tan{(x + a) - tan(x + b)] dx

= {1/sin(a - b)} * [-log|cos(x + a)| + log|cos(x + b)|] + C

= {1/sin(a - b)} * log|cos(x + b) /cos(x + a)| + C

Question 12:

x3/√(1 – x8)

Given, x3/√(1 – x8) = x3/√{1 – (x4)2}

Let x4 = t

=> 4x3 dx = dt

=> x3 dx = dt/4

So, ꭍ [x3/√(1 – x8)] = (1/4) * ꭍ dt/√(1 – t2)

= (1/4) * sin-1 t + C

= (1/4) * sin-1 (x4) + C

Question 13:

ex/{(1 + ex)(2 + ex)}

Let ex = t

=> ex dx = dt

So, ꭍ [ex/{(1 + ex)(2 + ex)}] dx = ꭍ dt/{(t + 1)(t + 2)}

= ꭍ [1/(t + 1) – 1/(t + 2)]dt

= log|t + 1| – log|t + 1| + C

= log|(t + 1)/(t + 2)| + C

= log|(1 + t)/(2 + t)| + C

= log|(1 + ex)/(2 + ex)| + C

Question 14:

1/{(x2 + 1)(x2 + 4)}

Let 1/{(x2 + 1)(x2 + 4)} = (Ax + B)/(x2 + 1) + (Cx + D)/(x2 + 4)

=> 1 = (Ax + B)(x2 + 4) + (Cx + D)(x2 + 1)

=> 1 = Ax3 + 4Ax + Bx2 + 4B + Cx3 + 4Cx + Dx2 + D

Equating the coefficients of x3, x2, x, and constant term, we get

A + C = 0

B + D = 0

4A + C = 0

4B + D = 1

On solving these equations, we get

A = 0, B = 1/3, C = 0 and D = -1/3

From equation 1, we get

1/{(x2 + 1)(x2 + 4)} = 1/3(x2 + 1) - 1/3(x2 + 4)

Now, ꭍ dx/{(x2 + 1)(x2 + 4)} = (1/3) * ꭍ dx/(x2 + 1) – (1/3) * ꭍ dx/(x2 + 4)

= (1/3) * tan-1 (x/1) – (1/3) * (1/2) * tan-1 (x/2) + C

= (1/3) * tan-1 (x) – (1/6) * tan-1 (x/2) + C

Question 15:

cos3 x * elog sin x

cos3 x * elog sin x = cos3 x * sin x

Let cos x = t

=> -sin x dx = dt

=> sin x dx = -dt

So, ꭍ cos3 x * elog sin x dx = ꭍ cos3 x * sin x dx

= -ꭍ t3 dt

= -t4/4 + C

= - cos4 x /4 + C

Question 16:

e3 log x * (x4 + 1)-1

e3 log x * (x4 + 1)-1 = e3 log x3 * 1/(x4 + 1)

= x3 * 1/(x4 + 1)

= x3/(x4 + 1)

Let x4 + 1 = t

=> 4x3 dx = dt

=> x3 dx = dt/4

So, ꭍ e3 log x * (x4 + 1)-1 dx = ꭍ [x3/(x4 + 1)] dx

= (1/4) * ꭍ dt/t

= (1/4) * log|t| + C

= (1/4) * log(x4 + 1) + C

Question 17:

f’(ax + b)[f(ax + b)]n

Let f(ax + b) = t

=> a * f’(ax + b) dx = dt

=> f’(ax + b) dx = dt/a

So, ꭍ f’(ax + b)[f(ax + b)]n dx = (1/a) * ꭍ tn dt

= (1/a) * tn+1/(n + 1) + C

= 1/{a(n + 1)} * [f(ax + b)]n+1 + C

Question 18:

1/√{sin3 x * sin(x + α)}

1/√{sin3 x * sin(x + α)} = 1/√{sin3 x * (sin x * cos α + cos x * sin α)}

= 1/√{sin4 x * cos α + sin3 x * cos x * sin α)}

= 1/[sin2 x√{cos α + cot x * sin α}]

= cosec2 x/√{cos α + cot x * sin α}

Let cos α + cot x * sin α = t

=> - cosec2 x * sin α dx = dt

=> - cosec2 x dx = -dt/sin α

So, ꭍ dx/√{sin3 x * sin(x + α)} = ꭍ [cosec2 x/√{cos α + cot x * sin α}] dx

= (-1/sin α) * ꭍ dt/√t

= (-2/sin α) * √t + C

= (-2/sin α) * √{cos α + cot x * sin α} + C

= (-2/sin α) * √{cos α + (cos x * sin α)/sin x} + C

= (-2/sin α) * √{(sin x * cos α + cos x * sin α)/sin x} + C

= (-2/sin α) * √{sin (x + α)/sin x} + C

Question 19:

(sin-1 √x - cos-1 √x)/(sin-1 √x + cos-1 √x), x є [0, 1]

Let I = ꭍ [(sin-1 √x - cos-1 √x)/(sin-1 √x + cos-1 √x)] dx

We know that sin-1 √x + cos-1 √x = π/2

So, I = ꭍ [(π/2 - cos-1 √x - cos-1 √x)/(π/2)] dx

=> I = (2/π) * ꭍ [π/2 – 2 * cos-1 √x] dx

=> I = (2/π) * ꭍ (π/2) dx - (2/π) * ꭍ 2 * cos-1 √x dx

=> I = ꭍ dx - (4/π) * ꭍ cos-1 √x dx

=> I = x - (4/π) * ꭍ cos-1 √x dx

=> I = x - (4/π) * I1   …………1

Now, I1 = ꭍ cos-1 √x dx

Let √x = t

=> (1/2√x) dx = dt

=> dx = 2√x dt

=> dx = 2t dt

Now, I1 = 2 * ꭍ cos-1 t * t dt

= 2 * [cos-1 t * ꭍ t dt - ꭍ{d(cos-1 t)/dx * ꭍ t dt}dt]

= 2 * [cos-1 t * t2/2 - ꭍ{-1/√(1 – t2) * t2/2}dt]

= t2 * cos-1 t + ꭍ {t2/√(1 – t2)}dt

= t2 * cos-1 t - ꭍ {(1 - t2 -1)/√(1 – t2)}dt

= t2 * cos-1 t - ꭍ {(1 - t2)/√(1 – t2) - 1/√(1 – t2)}dt

= t2 * cos-1 t - ꭍ √(1 – t2) dt - ꭍ dt/√(1 – t2)

= t2 * cos-1 t – (t/2) * √(1 – t2) – (1/2) * sin-1 t + sin-1 t

= t2 * cos-1 t – (t/2) * √(1 – t2) + (1/2) * sin-1 t

From equation 1, we get

=> I = x - (4/π) * [t2 * cos-1 t – (t/2) * √(1 – t2) + (1/2) * sin-1 t] + C

=> I = x - (4/π) * [x * cos-1 √x – (√x/2) * √(1 – x) + (1/2) * sin-1 √x] + C

=> I = x - (4/π) * [x * (π/2 - sin-1 √x) – √(x – x2)/2 + (1/2) * sin-1 √x] + C

=> I = x - (4/π) * [πx/2 – x * sin-1 √x) – √(x – x2)/2 + (1/2) * sin-1 √x] + C

=> I = x - (4/π) * (πx/2) – (4/π) * (x * sin-1 √x) – (4/π) *√(x – x2)/2 + (4/π) * (1/2) * sin-1 √x + C

=> I = x – 2x + (4x/π) * sin-1 √x + (2/π) * √(x – x2) - (2/π) * sin-1 √x + C

=> I = -x + (2/π) * [(2x - 1) * sin-1 √x] + (2/π) * √(x – x2) + C

=> I = {2(2x - 1)/π} * sin-1 √x + (2/π) * √(x – x2) - x + C

So, ꭍ [(sin-1 √x - cos-1 √x)/(sin-1 √x + cos-1 √x)] dx = {2(2x - 1)/π} * sin-1 √x + (2/π) * √(x – x2) - x + C

Question 20:

√{(1 - √x)/(1 + √x)}

Let I = ꭍ [√{(1 - √x)/(1 + √x)}] dx

Again let x = cos2 θ

=> dx = -2 * cos θ * sin θ dθ

So, I = ꭍ [√{(1 - cos θ)/(1 + cos θ)} * (-2 * cos θ * sin θ)] dθ

=> I = -ꭍ [√{(2 * sin2 θ/2)/(2 * cos2 θ/2)} * sin 2θ] dθ

=> I = -ꭍ [(sin θ/2)/(cos θ/2) * sin 2θ] dθ

=> I = -ꭍ [(sin θ/2)/(cos θ/2) * (2 * sin θ * cos θ)] dθ

=> I = -2 * ꭍ [(sin θ/2)/(cos θ/2) * (2 * sin θ/2 * cos θ/2) * cos θ] dθ

=> I = -4 * ꭍ [sin2 θ/2 * cos θ] dθ

=> I = -4 * ꭍ [sin2 θ/2 * (2 * cos2 θ/2 - 1)] dθ

=> I = -4 * ꭍ [2 * sin2 θ/2 * cos2 θ/2 - sin2 θ/2] dθ

=> I = -8 * ꭍ sin2 θ/2 * cos2 θ/2 dθ + 4 * ꭍ sin2 θ/2 dθ

=> I = -2 * ꭍ (4 * sin2 θ/2 * cos2 θ/2) dθ + 4 * ꭍ sin2 θ/2 dθ

=> I = -2 * ꭍ (2 * sin θ/2 * cos θ/2)2 dθ + 4 * ꭍ sin2 θ/2 dθ

=> I = -2 * ꭍ (sin θ)2 dθ + 4 * ꭍ (1 - cos θ)/2 dθ

=> I = -2 * ꭍ sin2 θ dθ + 4 * ꭍ (1 - cos θ)/2 dθ

=> I = -2 * ꭍ (1 - cos 2θ)/2 dθ + 4 * ꭍ (1 - cos θ)/2 dθ

=> I = -ꭍ (1 - cos 2θ) dθ + 2 * ꭍ (1 - cos θ) dθ

=> I = -(θ - sin 2θ /2) + 2 * (θ - sin θ) + C

=> I = -θ + sin 2θ /2 + 2θ – 2 * sin θ + C

=> I = θ + (2 * sin θ * cos θ)/2 – 2 * sin θ + C

=> I = θ + sin θ * cos θ – 2 * sin θ + C

=> I = θ + √(1 – cos2 θ) * cos θ – 2 * √(1 – cos2 θ) + C

=> I = cos-1 √x + √(1 – x) * √x – 2 * √(1 – x) + C

=> I = -2√(1 – x) + cos-1 √x + √(x – x2) + C

So, √{(1 - √x)/(1 + √x)} = -2√(1 – x) + cos-1 √x + √(x – x2) + C

Question 21:

[(2 + sin 2x)/(1 + cos 2x)] * ex

Let I = ꭍ [(2 + sin 2x)/(1 + cos 2x)] * ex dx

= ꭍ [(2 + 2 * sin x * cos x)/(2 cos2 x)] * ex dx

= ꭍ [(1 + sin x * cos x)/cos2 x] * ex dx

= ꭍ (sec2 x + tan x)/ * ex dx

Let f(x) = tan x

=> f’(x) = sec2 x

So, I = ꭍ [f(x) + f’(x)] * ex dx

=> I = ex * f(x) + C

=> I = ex * tan x + C

=> ꭍ [(2 + sin 2x)/(1 + cos 2x)] * ex dx = ex * tan x + C

Question 22:

(x2 + x + 1)/{(x + 1)2(x + 2)}

Let (x2 + x + 1)/{(x + 1)2(x + 2)} = A/(x + 1) + B/(x + 1)2 + C/(x + 2)     ………….1

=> x2 + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)2

=> x2 + x + 1 = A(x2 + 2x + 1) + Bx + 2B + C(x2 + 2x + 1)

=> x2 + x + 1 = (A + C)x2 + (3A + B + 2C)x + (2A + 2B + C)

Equating the coefficients of x2, x and constant term, we get

A + C = 1

3A + B + 2C = 1

2A + 2B + C = 1

On solving these equations, we get

A = −2, B = 1, and C = 3

From equation 1, we get

(x2 + x + 1)/{(x + 1)2(x + 2)} = -2/(x + 1) + 1/(x + 1)2 + 3/(x + 2)

Now, ꭍ [(x2 + x + 1)/{(x + 1)2(x + 2)}] dx = -2 * ꭍ dx/(x + 1) + ꭍ dx/(x + 1)2 + 3 * ꭍ dx/(x + 2)

= -2 * log|x + 1| - 1/(x + 1) + 3 * log|x + 2| + C

Question 23:

tan-1 √{(1 - x)/(1 + x)}

Let I = ꭍ tan-1 √{(1 - x)/(1 + x)} dx

Again let x = cos θ

=> dx = -sin θ dθ

Now, I = ꭍ [tan-1 √{(1 – cos θ)/(1 + cos θ)} * (-sin θ )] dθ

= - ꭍ [tan-1 √{(2 * sin2 θ/2)/(2 * cos2 θ/2)} * (sin θ)] dθ

= - ꭍ [tan-1 √(tan2 θ/2) * (sin θ)] dθ

= - ꭍ [tan-1(tan2 θ/2) * (sin θ)] dθ

= - ꭍ [(θ/2) * (sin θ)] dθ

= (-1/2) * ꭍ θ * sin θ dθ

= (-1/2) * [θ * ꭍ sin θ dθ - ꭍ {d(θ)/dθ * ꭍ sin θ dθ} dθ]

= (-1/2) * [θ * (-cos θ) - ꭍ (-cos θ) dθ]

= (-1/2) * [-θ * cos θ + sin θ] + C

= (1/2) * (θ * cos θ) - (1/2) * sin θ + C

= (1/2) * (cos-1 x * x) - (1/2) * √(1 – x2) + C

= (x/2) * cos-1 x - (1/2) * √(1 – x2) + C

= (1/2) * [x * cos-1 x - √(1 – x2)] + C

So, tan-1 √{(1 - x)/(1 + x)} = (1/2) * [x * cos-1 x - √(1 – x2)] + C

Question 24:

[√(x2 + 1){log(x2 + 1) – 2 * log x}]/x4

Let I = ꭍ [√(x2 + 1){log(x2 + 1) – 2 * log x}]/x4 dx     ………….1

Now, [√(x2 + 1){log(x2 + 1) – 2 * log x}]/x4 = [√(x2 + 1)/x4] * {log(x2 + 1) – 2 * log x}

= [√(x2 + 1)/x4] * {log(x2 + 1) – log x2}

= [√(x2 + 1)/x4] * [log{(x2 + 1)/x2}]

= [√(x2 + 1)/x4] * [log(1 + 1/x2)]

= [√{(x2 + 1)/x2} * (1/x3)] * [log(1 + 1/x2)]

= [√(1 + 1/x2) * (1/x3)] * [log(1 + 1/x2)]

Let 1 + 1/x2 = t

=> (-2/x3) dx = dt

=> (1/x3) dx = dt/(-2)

So, I = (-1/2) * ꭍ √t * log t dt

=> I = (-1/2) * ꭍ t1/2 * log t dt

Integrating by parts, we get

=> I = (-1/2) * [log t * ꭍ t1/2 dt - ꭍ {d(log t)/dt * ꭍ t1/2 dt}dt]

=> I = (-1/2) * [log t * t3/2/(3/2) - ꭍ {1/t * t3/2/(3/2)}dt]

=> I = (-1/2) * [(2/3) * t3/2 * log t - (2/3) * ꭍ t1/2 dt]

=> I = (-1/2) * [(2/3) * t3/2 * log t - (2/3) * t3/2/(3/2)] + C

=> I = (-1/2) * [(2/3) * t3/2 * log t - (4/9) * t3/2] + C

=> I = (-1/3) * t3/2 * log t + (2/9) * t3/2 + C

=> I = (-1/3) * t3/2 * [log t – 2/3] + C

=> I = (-1/3) * (1 + 1/x2)3/2 * [log(1 + 1/x2) – 2/3] + C

Evaluate the definite integrals in Exercises 25 to 33.

Question 25:

π/2π ex {(1 – sin x)/(1 + sin x)} dx

Let I = ꭍπ/2π ex {(1 – sin x)/(1 + sin x)} dx

= ꭍπ/2π ex {(1 – 2 * sin x/2 * cos x/2)/(2 * sin2 x/2)} dx

= ꭍπ/2π ex {1/(2 * sin2 x/2) – (2 * sin x/2 * cos x/2)/(2 * sin2 x/2)} dx

= ꭍπ/2π ex {(cosec2 x/2)/2 – cot x/2} dx

Let f(x) = -cot x/2

=> f’(x) = -{(-1/2) * cosec2 x/2}

=> f’(x) = (1/2) * cosec2 x/2

So, I = ꭍπ/2π ex {f(x) + f’(x)} dx

= [ex * f(x)]π/2π

= -[ex * cot x/2]π/2π

= -[eπ * cot π/2 - eπ/2 * cot π/4]

= -[eπ * 0 - eπ/2 * 1]

= eπ/2

So, ꭍπ/2π ex {(1 – sin x)/(1 + sin x)} dx = eπ/2

Question 26:

0 π/4 [(sin x * cos x)/(cos4 x + sin4 x)] dx

Let I = ꭍ0 π/4 [(sin x * cos x)/(cos4 x + sin4 x)] dx

=> I = ꭍ0 π/4 [{(sin x * cos x)/cos4 x}/{(cos4 x + sin4 x)/cos4 x}] dx

=> I = ꭍ0 π/4 [(tan x * sec2 x)/(1 + tan4 x)] dx

=> I = ꭍ0 π/4 [(tan x * sec2 x)/{1 + (tan2 x)2}] dx

Let tan2 x = t

=> 2 * tan x * sec2 x dx = dt

=> tan x * sec2 x dx = dt/2

When x = 0, t = 0 and when x = π/4, t = 1

So, I = (1/2) * ꭍ01 dt/(1 + t2)

=> I = (1/2) * [tan-1 t]01

=> I = (1/2) * [tan-1 1 - tan-1 0]

=> I = (1/2) * (π/4)

=> I = π/8

=> ꭍ0 π/4 [(sin x * cos x)/(cos4 x + sin4 x)] dx = π/8

Question 27:

0π/2 [cos2 x/(cos2 x + 4 * sin2 x)] dx

Let I = ꭍ0π/2 [cos2 x/(cos2 x + 4 * sin2 x)] dx

=> I = ꭍ0π/2 [cos2 x/{cos2 x + 4(1 - cos2 x)}] dx

=> I = ꭍ0π/2 [cos2 x/(cos2 x + 4 – 4 * cos2 x)] dx

=> I = (-1/3) * ꭍ0π/2 [(-3 * cos2 x)/(4 – 3 * cos2 x)] dx

=> I = (-1/3) * ꭍ0π/2 [(4 - 3 * cos2 x - 4)/(4 – 3 * cos2 x)] dx

=> I = (-1/3) * ꭍ0π/2 [(4 - 3 * cos2 x)/(4 – 3 * cos2 x)] dx + (1/3) * ꭍ0π/2 [4/(4 – 3 * cos2 x)] dx

=> I = (-1/3) * ꭍ0π/2 dx + (1/3) * ꭍ0π/2 [(4 * sec2 x)/(4 * sec2 x – 3)] dx

=> I = (-1/3) * [x]0π/2 + (1/3) * ꭍ0π/2 [(4 * sec2 x)/{4 * (1 + tan2 x) – 3}] dx

=> I = (-1/3) * [π/2] + (1/3) * ꭍ0π/2 [(4 * sec2 x)/(4 + 4 * tan2 x – 3)] dx

=> I = -π/6 + (2/3) * ꭍ0π/2 [(2 * sec2 x)/(1 + 4 * tan2 x)] dx

=> I = -π/6 + (2/3) * ꭍ0π/2 [(2 * sec2 x)/{1 + (2 * tan x)2}] dx      …………..1

Consider, ꭍ0π/2 [(2 * sec2 x)/{1 + (2 * tan x)2}] dx

Let 2 * tan x = t

=> 2 * sec2 x dx = dt

When x = 0, t = 0 ans when x = π/2, t = ∞

So, ꭍ0π/2 [(2 * sec2 x)/{1 + (2 * tan x)2}] dx = ꭍ0 dt/(1 + t2)

= [tan-1 t]0

= (tan-1 ∞ - tan-1 0)

= π/2

From equation 1, we get

=> I = -π/6 + (2/3) * (π/2)

=> I = -π/6 + π/3

=> I = π/6

So, ꭍ0π/2 [cos2 x/(cos2 x + 4 * sin2 x)] dx = π/6

Question 28:

π/6π/3 [(sin x + cos x)/√(sin 2x)] dx

Let I = ꭍπ/6π/3 [(sin x + cos x)/√(sin 2x)] dx

=> I = ꭍπ/6π/3 [(sin x + cos x)/{-√(-sin 2x)}] dx

=> I = ꭍπ/6π/3 [(sin x + cos x)/{-√(-1 + 1 + 2 * sin x * cos x)}] dx

=> I = ꭍπ/6π/3 [(sin x + cos x)/√(1 - 1 - 2 * sin x * cos x)] dx

=> I = ꭍπ/6π/3 [(sin x + cos x)/√{1 – (sin2 x + cos2 x - 2 * sin x * cos x)}] dx

=> I = ꭍπ/6π/3 [(sin x + cos x)/√{1 – (sin x – cos x)2}] dx

Let sin x – cos x = t

=> (sin x + cos x)dx = dt

When x = π/6, t = (1 - √3)/2 and when x = π/3, t = (√3 - 1)/2

Now, I = ꭍ(1 - √3)/2(√3 - 1)/2 dt/√(1 – t2)

=> I = ꭍ-(√3 - 1)/2(√3 - 1)/2 dt/√(1 – t2)

Since ꭍ dt/√{1 – (-t)2} = dt/√(1 – t2)

So, dt/√(1 – t2) is an even function.

It is known that if f(x) is an even function, then

-aa f(x) dx = 2 * ꭍ0a f(x) dx

Now, I = 2 * ꭍ0(√3 - 1)/2 dt/√(1 – t2)

=> I = 2 * [sin-1 t]0(√3 - 1)/2

=> I = 2 * [sin-1 {(√3 - 1)/2} - sin-1 0]

=> I = 2 * sin-1 {(√3 - 1)/2}

Hence, ꭍπ/6π/3 [(sin x + cos x)/√(sin 2x)] dx = 2 * sin-1 {(√3 - 1)/2}

Question 29:

01 dx/{√(1 + x) - √x}

Let I = ꭍ01 dx/{√(1 + x) - √x}

=> I = ꭍ01 [1 * {√(1 + x) + √x}]/[{√(1 + x) - √x} * {√(1 + x) + √x}] dx

=> I = ꭍ01 [{√(1 + x) + √x}/(1 + x – x)] dx

=> I = ꭍ01 [√(1 + x) + √x] dx

=> I = ꭍ01 (1 + x)1/2 dx + ꭍ01 x1/2 dx

=> I = (2/3) * [(1 + x)3/2]01 + (2/3) * [x3/2]01

=> I = (2/3) * [(1 + 1)3/2 – 1] + (2/3) * 13/2

=> I = (2/3) * [23/2 - 1] + (2/3) * 1

=> I = (2/3) * [23/2 – 1 + 1]

=> I = (2/3) * 23/2

=> I = (2/3) * (2√2)

=> I = 4√2/3

=> ꭍ01 dx/{√(1 + x) - √x} = 4√2/3

Question 30:

0π/4 [(sin x + cos x)/(9 + 16 * sin 2x)]dx

Let I = ꭍ0π/4 [(sin x + cos x)/(9 + 16 * sin 2x)]dx

Also let sin x – cos x = t

=> (cos x + sin x)dx = dt

When x = 0, t = -1 and when x = π/4, t = 0

Again (sin x – cos x)2 = t2

=> sin2 x + cos2 x - 2 * sin x * cos x = t2

=> 1 - sin 2x = t2

=> sin 2x = 1 - t2

So,I =  ꭍ-10 dt/{9 + 16(1 - t2)} dx

=> I = ꭍ-10 dt/(9 + 16 - 16t2) dx

=> I = ꭍ-10 dt/(25 - 16t2) dx

=> I = ꭍ-10 dt/{52 – (4t)2} dx

=> I = (1/4) * [1/(2 * 5) * log|(5 + 4t)/(5 - 4t)|]-10

=> I = (1/40) * [log (1) – log (1/9)]

=> I = (1/40) * [0 + log 9]

=> I = (1/40) * log 9

So, ꭍ0π/4 [(sin x + cos x)/(9 + 16 * sin 2x)]dx = (1/40) * log 9

Question 31:

0π/2 [sin 2x * tan-1(sin x)] dx

Let I = ꭍ0π/2 [sin 2x * tan-1(sin x)] dx

Let I = ꭍ0π/2 [2 * sin x * cos x * tan-1(sin x)] dx

Let I = 2 * ꭍ0π/2 [sin x * cos x * tan-1(sin x)] dx

Let sin x = t

=> cos x dx = dt

When x = 0, t = 0 and when x = π/2, t = 1

Now, I = 2 * ꭍ01 t * tan-1 t dt    ……………1

Consider ꭍ t * tan-1 t dt = tan-1 t * ꭍ t * dt - ꭍ {d(tan-1 t)/dt * ꭍ t * dt}dt

= tan-1 t * t2/2 - ꭍ {1/(1 + t2) * (t2/2)}dt

= (t2 * tan-1 t)/2 – (1/2) * ꭍ [t2/(1 + t2)] dt

= (t2 * tan-1 t)/2 – (1/2) * ꭍ [(t2 + 1 – 1)/(1 + t2)] dt

= (t2 * tan-1 t)/2 – (1/2) * ꭍ [(t2 + 1)/(1 + t2)] dt + (1/2) * ꭍ dt/(1 + t2)]

= (t2 * tan-1 t)/2 – (1/2) * ꭍ dt + (1/2) * ꭍ dt/(1 + t2)]

= (t2 * tan-1 t)/2 – t/2 + (1/2) * tan-1 t

Now, ꭍ01 t * tan-1 t dt = [(t2 * tan-1 t)/2 – t/2 + (1/2) * tan-1 t]01

= [(12 * tan-1 1)/2 – 1/2 + (1/2) * tan-1 1 - 0]

= (π/4)/2 – 1/2 + (1/2) * (π/4)

= (1/2) * (π/4 – 1 + π/4)

= (1/2) * (π/2 – 1)

= π/4 – 1/2

From equation 1, we get

=> I = 2 * (π/4 – 1/2)

=> I = π/2 – 1

Hence, ꭍ0π/2 [sin 2x * tan-1(sin x)] dx = π/2 – 1

Queston 32:

0π [(x * tan x)/(sec x + tan x)] dx

Let I = ꭍ0π [(x * tan x)/(sec x + tan x)] dx    …………..1

=> I = ꭍ0π [{(π - x) * tan (π - x)}/{sec (π - x) + tan (π - x)}] dx

=> I = ꭍ0π [{-(π - x) * tan x}/{-(sec x + tan x)}] dx

=> I = ꭍ0π [{(π - x) * tan x}/(sec x + tan x)] dx     …………2

Adding equations 1 and 2, we get

=> 2I = ꭍ0π [(π * tan x)/(sec x + tan x)] dx

=> 2I = π * ꭍ0π [tan x/(sec x + tan x)] dx

=> 2I = π * ꭍ0π [(sin x /cos x)/(1/cos x + sin x /cos x)] dx

=> 2I = π * ꭍ0π [sin x /(1 + sin x)] dx

=> 2I = π * ꭍ0π [(1 + sin x - 1)/(1 + sin x)] dx

=> 2I = π * ꭍ0π [(1 + sin x)/(1 + sin x)] dx - π * ꭍ0π dx/(1 + sin x)

=> 2I = π * ꭍ0π dx - π * ꭍ0π dx/(1 + sin x)

=> 2I = π * [x]0π - π * ꭍ0π [(1 – sin x)/{(1 + sin x)(1 - sin x)}] dx

=> 2I = π2 - π * ꭍ0π [(1 – sin x)/(1 – sin2 x)] dx

=> 2I = π2 - π * ꭍ0π [(1 – sin x)/cos2 x] dx

=> 2I = π2 - π * ꭍ0π [1/cos2 x – sin x/cos2 x] dx

=> 2I = π2 - π * ꭍ0π [sec2 x – tan x * sec x] dx

=> 2I = π2 - π * [tan x –sec x]0π

=> 2I = π2 - π * [(tan π – sec π) - (tan 0 – sec 0)]

=> 2I = π2 - π * [(0 + 1) - (0 – 1)]

=> 2I = π2 - 2π

=> 2I = π(π – 2)

=> I = π(π – 2)/2

Hence, ꭍ0π [(x * tan x)/(sec x + tan x)] dx = π(π – 2)/2

Question 33:

14 [|x - 1| + |x - 2| + |x - 3|] dx

Let I = ꭍ14 [|x - 1| + |x - 2| + |x - 3|] dx

=> I = ꭍ14 |x - 1| dx + ꭍ14|x - 2| dx + ꭍ14|x - 3| dx

=> I = I1 + I2 + I3

Where I1 = ꭍ14 |x - 1| dx, I2 = ꭍ14|x - 2| dx, I3 = ꭍ14|x - 3| dx

Now, I1 = ꭍ14 |x - 1| dx

Since (x - 1) ≥ 0 for 1 ≤ x ≤ 4

So, I1 = ꭍ14 (x – 1) dx

=> I1 = [x2/2 – x]14

=> I1 = [42/2 – 4] - [12/2 – 1]

=> I1 = (8 – 4) – (1/2 - 1)

=> I1 = 4 + 1/2

=> I1 = 9/2     …………2

I2 = ꭍ14|x - 2| dx

Since (x - 2) ≥ 0 for 2 ≤ x ≤ 4 and (x - 2) ≤ 0 for 1 ≤ x ≤ 2

So, I2 = -ꭍ12 (x – 2) dx + ꭍ24(x – 2) dx

=> I2 = ꭍ12 (2 – x) dx + ꭍ24(x – 2) dx

=> I2 = [2x - x2/2]12 + [x2/2 – 2x]24

=> I2 = [(4 – 2) – (2 – 1/2)] + [(8 - 8) – (2 - 4)]

=> I2 = [4 – 2 – 2 + 1/2] + [8 - 8 – 2 + 4]

=> I2 = 1/2 + 2

=> I2 = 5/2      ………….3

I3 = ꭍ14|x - 3| dx

Since (x - 3) ≥ 0 for 3 ≤ x ≤ 4 and (x - 3) ≤ 0 for 1 ≤ x ≤ 3

So, I3 = -ꭍ13 (x – 3) dx + ꭍ34(x – 3) dx

=> I3 = ꭍ13 (3 – x) dx + ꭍ34(x – 3) dx

=> I3 = [3x - x2/2]13 + [x2/2 – 3x]34

=> I3 = [(9 – 9/2) – (3 – 1/2)] + [(8 - 12) – (9/2 - 9)]

=> I3 = [9 – 9/2 – 3 + 1/2] + [8 - 12 – 9/2 + 9]

=> I3 = 6 - 4 + 1/2

=> I3 = 5/2      ………….4

From equations 1, 2, 3 and 4, we get

=> I = 9/2 + 5/2 + 5/2

=> I = (9 + 5 + 5)/2

=> I = 19/2

Hence, ꭍ14 [|x - 1| + |x - 2| + |x - 3|] dx = 19/2

Prove the following (Exercises 34 to 39)

Question 34:

13 dx/{x2(x + 1)} = 2/3 + log (2/3)

Let I = ꭍ13 dx/{x2(x + 1)}     ………….1

Again let 1/{x2(x + 1)} = A/x + B/x2 + C/(x + 1)      ………….2

=> 1 = Ax(x + 1) + B(x + 1) + Cx2

=> 1 = Ax2 + Ax + Bx + B + Cx2

Equating the coefficients of x2, x, and constant term, we get

A + C = 0

A + B = 0

B = 1

On solving these equations, we obtain

A = −1, C = 1, and B = 1

From equation 2, we get

=> 1/{x2(x + 1)} = -1/x + 1/x2 + 1/(x + 1)

From equation 1, we get

=> I = ꭍ13 dx/{x2(x + 1)}

=> I = ꭍ13 [-1/x + 1/x2 + 1/(x + 1)]

=> I = [-log x – 1/x + log(x + 1)]13

=> I = [log {(x + 1)/x} – 1/x]13

=> I = [log {(3 + 1)/3} – 1/3] - [log {(1 + 1)/1} – 1/1]

=> I = log (4/3) – 1/3 - [log 2 – 1]

=> I = log (4/3) – 1/3 - log 2 + 1

=> I = log 4 – log 3 – 1/3 - log 2 + 1

=> I = log 22 – log 3 – 1/3 - log 2 + 1

=> I = 2 * log 2 – log 3 – 1/3 - log 2 + 1

=> I = log 2 – log 3 + 2/3

=> I = log (2/3) + 2/3

Hence, the given result is proved.

Question 35:

01 x * ex dx = 1

Let I = ꭍ01 x * ex dx

Integrating by parts, we get

=> I = x * ꭍ01 ex dx - ꭍ01 [d(x)/dx * ꭍ ex dx]dx

=> I = [xex]01 - ꭍ01 ex dx

=> I = [xex]01 - [ex]01

=> I = [1 * e1 – 0] - [e1 – e0]

=> I = e – e + 1

=> I = 1

=> ꭍ01 x * ex dx = 1

Hence, the given result is proved.

Question 36:

-11 x17 * cos4 x dx = 0

Let I = ꭍ-11 x17 * cos4 x dx

Also let f(x) = x17 * cos4 x

Now, f(-x) = (-x)17 * cos4 (-x) = - x17 * cos4 x = -f(x)

Therefore, f(x) is an odd number.

It is known that if f(x) is an odd function, then ꭍ-aa f(x) dx = 0

So, I = ꭍ-11 x17 * cos4 x dx = 0

Hence, the given result is proved.

Question 37:

0π/2 sin3 x dx = 2/3

Let I = ꭍ0π/2 sin3 x dx

=> I = ꭍ0π/2 sin2 x * sin x dx

=> I = ꭍ0π/2 (1 - cos2 x) * sin x dx

=> I = ꭍ0π/2 sin x dx - ꭍ0π/2 cos2 x * sin x dx

=> I = -[cos x]0π/2 + [cos3 x /3]0π/2

=> I = -[cos π/2 – cos 0] + (1/3) * [cos3 π/2 - cos3 0]

=> I = -[0 – 1] + (1/3) * [0 - 1]

=> I = 1 - 1/3

=> I = 2/3

=> ꭍ0π/2 sin3 x dx = 2/3

Hence, the given result is proved.

Question 38:

0π/4 2 tan3 x dx = 1 – log 2

Let I = ꭍ0π/4 2 tan3 x dx

=> I = 2 * ꭍ0π/4 tan2 x * tan x dx

=> I = 2 * ꭍ0π/4 (sec2 x – 1) * tan x dx

=> I = 2 * ꭍ0π/4 sec2 x * tan x dx – 2 * ꭍ0π/4 tan x dx

=> I = 2 * [tan2 x /2]0π/4 + 2 * [log cos x]0π/4

=> I = [tan2 π/4 - tan2 0] + 2 * [log cos π/4 – log cos 0]

=> I = 1 + 2 * [log (1/√2) – log 1]

=> I = 1 + (2/2) * log (1/2)

=> I = 1 - log 2

=> ꭍ0π/4 2 tan3 x dx = 1 – log 2

Hence, the given result is proved.

Question 39:

01 sin-1 x dx = π/2 – 1

Let I = ꭍ01 sin-1 x dx

=> I = ꭍ01 sin-1 x * 1 dx

Integrating by parts, we get

=> I = sin-1 x * ꭍ01 1 dx - ꭍ[d(sin-1 x)/dx * ꭍ01 1 dx] dx

=> I = [x * sin-1 x]01 - ꭍ01 [x/√(1 – x2)] dx

=> I = [x * sin-1 x]01 – (1/2) * ꭍ01 [(-2x)/√(1 – x2)] dx

Let 1 – x2 = t

=> -2x dx = dt

When x = 0, t = 1 and x = 1, t = 0

So, I = [x * sin-1 x]01 + (1/2) * ꭍ10 dt/√t

=> I = [x * sin-1 x]01 + (1/2) *  [2/√t]10

=> I = [sin-1 1] + [0 - √1]

=> I = π/2 – 1

=> ꭍ01 sin-1 x dx = π/2 – 1

Hence, the given result is proved.

Question 40:

Evaluate ꭍ01 e2 – 3x dx as a limit of a sum.

Let I = ꭍ01 e2 – 3x dx

We know that

ab f(x) dx = (b - a) * limn->∞ (1/n)[f(a) + f(a + h) + ………..+ f{a + (n - 1)h}]

Where h = (b - a)/n

Here a = 0, b = 1 and f(x) = e2 – 3x

So, h = (1 - 0)/n = 1/n

Now, ꭍ01 e2 – 3x dx = (1 - 0) * limn->∞ (1/n)[f(0) + f(0 + h) + ………..+ f{0 + (n - 1)h}]

= limn->∞ (1/n)[e2 + e2-3h + ………..+ e2-3(n - 1)h]

= limn->∞ (1/n)[e2{1 + e-3h + e-9h +………..+ e-3(n - 1)h]

= limn->∞ (1/n)[e2{(1 – (e-3h)n)/(1 – e-3h)}]

= limn->∞ (1/n)[e2{(1 – e-3/n *n)/(1 – e-3/n)}]

= limn->∞ (1/n)[e2{(1 – e-3)/(1 – e-3/n)}]

= e2(e-3 – 1) * limn->∞ (1/n)[1/(e-3/n – 1)}]

= e2(e-3 – 1) * limn->∞ (-1/3)[(-3/n)/(e-3/n – 1)}]

= -e2(e-3 – 1)/3 * limn->∞ [(-3/n)/(e-3/n – 1)}]

= -e2(e-3 – 1)/3 * 1                                   [Since limn->∞ {x/(ex - 1)} = 1]

= -e2(e-3 – 1)/3

= (-e-1 + e2)/3

= (e2 – 1/e)/3

So, ꭍ01 e2 – 3x dx = (e2 – 1/e)/3

Choose the correct answers in Exercises 41 to 44.

Question 41:

ꭍ dx/(ex + e-x) is equal to

1. tan-1 (ex) + C B. tan-1 (e-x) + C               C. log(ex - e-x) + C             D. log(ex + e-x) + C

Let I = ꭍ dx/(ex + e-x)

=> I = ꭍ [ex/(e2x +1)] dx

Also, let ex = t

=> ex dx = dt

So, I = ꭍ dt/(1 + t2)

=> I = tan-1 t + C

=> I = tan-1 (ex) + C

=> ꭍ dx/(ex + e-x) = tan-1 (ex) + C

Hence, the correct answer is option A.

Question 42:

ꭍ [cos 2x /(sin x + cos x)2] dx is equal to

1. -1/(sin x + cos x) + C B. log|sin x + cos x| + C
2. log|sin x - cos x| + C D. 1/(sin x + cos x) + C

Let I = ꭍ [cos 2x /(sin x + cos x)2] dx

=> I = ꭍ [(cos2 x – sin2 x)/(sin x + cos x)2] dx

=> I = ꭍ [{(cos x – sin x)(cos x + sin x)}/(sin x + cos x)2] dx

=> I = ꭍ [(cos x – sin x)/(sin x + cos x)] dx

Let sin x + cos x = t

=> (-cos x + sin x) dx = dt

=> (sin x - cos x) dx = dt

So, I = ꭍ dt/t

=> I = log|t| + C

=> I = log|sin x + cos x| + C

=> ꭍ [cos 2x /(sin x + cos x)2] dx = log|sin x + cos x| + C

Hence, the correct answer is option B.

Question 43:

If f(a + b - x) = f(x), then ꭍab x * f(x) dx is equal to

1. (a + b)/2 * ꭍab f(b - x) dx B. (a + b)/2 * ꭍab f(b + x) dx
2. (b - a)/2 * ꭍab f(x) dx D. (a + b)/2 * ꭍab f(x) dx

Let I = ꭍab x * f(x) dx      …………….1

=> I = ꭍab (a + b - x) * f(a + b - x) dx                              [Since ꭍab f(x) dx = ꭍab (a + b - x) dx]

=> I = ꭍab (a + b - x) * f(x) dx

=> I = ꭍab (a + b) * f(x) dx - ꭍab x * f(x) dx

=> I = (a + b) * ꭍab f(x) dx – I                                      [Using equation 1]

=> 2I = (a + b) * ꭍab f(x) dx

=> I = (a + b)/2 * ꭍab f(x) dx

Hence, the correct answer is option D.

Question 44:

The value of ꭍ01 tan-1{(2x - 1)/(1 + x – x2)} dx is

1. 1 B. 0 C. -1                                           D. π/4

Let I = ꭍ01 tan-1{(2x - 1)/(1 + x – x2)} dx

=> I = ꭍ01 tan-1{(x – (1 - x))/(1 + x(1 - x))} dx

=> I = ꭍ01 [tan-1 x – tan-1 (1 - x)] dx      ………….1

=> I = ꭍ01 [tan-1 (1 - x) – tan-1 (1 – 1 + x)] dx                    [Since ꭍab f(x) dx = ꭍab (a + b - x) dx]

=> I = ꭍ01 [tan-1 (1 - x) – tan-1 (x)] dx    ………….2

Adding equations 1 and 2, we get

=> 2I = ꭍ01 [{tan-1 x – tan-1 (1 - x)} - {tan-1 (1 - x) – tan-1 (x)}] dx

=> 2I = ꭍ01 [tan-1 x – tan-1 (1 - x) - tan-1 (1 - x) + tan-1 (x)] dx

=> 2I = 0

=> I = 0

=> ꭍ01 tan-1{(2x - 1)/(1 + x – x2)} dx = 0

Hence, the correct answer is option B.