INTEGRATION PROBLEMS WORKSHEET

Problem 1 :

∫ (sec √x / √x) dx

Solution :

∫ (sec √x / √x) dx

Let us use substitution method to find the antiderivative of the given.

Let √x = t

Differentiate with respect to x, we get

(1/2 √x) dx = dt

(1/√x) dx = 2dt

∫ (sec √x) (1 / √x) dx = ∫ (sec t)(2 dt)

= 2∫ sec t dt

= 2 log (sec t + tan t) + C

= 2 log (sec √x + tan √x ) + C

Problem 2 :

∫ sin⁴ x dx

Solution :

∫ sin⁴ x dx = ∫ (sin² x)² dx

= ∫ [(1 - cos 2x)/2]² dx

= (1/4) ∫ (1 - 2cos 2x + cos² 2x) dx

From cos 2x = 2cos² x - 1

 cos² x = 1 + cos 2x

 cos² 2x = 1 + cos 4x

= (1/4) ∫ (1 - 2cos 2x + (1 + cos 4x)/2] dx

= (1/4) [∫ 1 dx - 2∫cos 2x dx + (1/2)∫(1 + cos 4x) dx]

= (1/4) [x - (2/2) sin 2x + (1/2)(x + (1/4)cos 4x)] + C

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

= (1/4) x - (1/4) sin 2x + (1/8)x + (1/32)cos 4x + C

= (3/8) x - (1/4) sin 2x + (1/32)cos 4x + C

Problem 3 :

∫ cos 3x dx

Solution :

∫ cos 3x dx = (1/3) sin (3x) + C

Problem 4 :

∫ sin 10x sin 4x dx

Solution :

∫ sin 10x sin 4x dx

- 2 sin C sin D = cos (C + D) - cos (C - D)

= (1/2) [cos (C - D) - cos (C + D)]

Here C = 10 and D = 4

= (1/2) ∫ [cos (10x - 4x) - cos (10x + 4x)] dx

= (1/2) ∫ [cos 6x - cos 14x] dx

= (1/2) [(1/6) sin 6x - (1/14) sin 14x] + C

= [(1/12) sin 6x - (1/28) sin 14x] + C

Problem 5 :

∫ sin 7x cos 3x dx

Solution :

∫ sin 7x cos 3x dx

sin (C + D) + sin (C - D) = 2sin C cos D

sin C cos D = (1/2) [sin (C + D) + sin (C - D)]

Here C = 7x and D = 3x

= (1/2) ∫ [sin (7x + 3x) + sin (7x - 3x)] dx

= (1/2) ∫ [sin 10x + sin 4x] dx

= (1/2) [(-1/10) cos 10x - (1/4) cos 4x] + C

= [(-1/20) cos 10x - (1/8) cos 4x] + C

Problem 6 :

Solution :

cos 2x = 1 - 2sin²x 

2sin²x = 1 - cos 2x

Dividing angle by 2 on both sides

2sin²(x/2) = 1 - cos (2x/2)

2sin²(x/2) = 1 - cos x

Problem 7 :

∫ sin³ x dx

Solution :

∫ sin³ x dx

sin 3x = 3 sin x - 4 sin³ x

4 sin³ x = 3 sin x - sin 3x

sin³ x = (1/4) [3 sin x - sin 3x]

∫ sin³ x dx = (1/4)∫ [3 sin x - sin 3x] dx

= (1/4) [-3 cos x + (1/3) cos 3x] + C

= [(-3/4) cos x + (1/12) cos 3x] + C

Problem 8 :

∫ tan³ x dx

Solution :

∫ tan³ x dx

= ∫ tan² x tan x dx ----(1)

Let t = tan x

dt = sec² x dx

Using trigonometric identity, let us apply the replacement of tan² x.

sec² x - tan² x = 1

tan² x = sec² x - 1

∫ tan² x tan x dx = ∫ (sec² x - 1) tan x dx 

= ∫(tan x sec² x - tan x) dx 

= ∫tan x sec² x dx - ∫ tan x dx 

= ∫t  dt - ∫ tan x dx 

 = t²/2 - log (cos x ) + C

Applying the value of t, we get

x = tan²/2 - log (cos x ) + C

Problem 9 :

∫(x² + 1) √(1+3x+x³) dx

Solution :

Let t = 1+3x+x³

dt = (3x² + 3)dx

dt = 3(x² + 1)dx

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