Integration Worksheet4 solution3

In this page integration worksheet4 solution3 we are going to see solution of some practice question from the worksheet of integration.

Question 8

Integrate the following with respect to x,cos² 2 x- sin 6 x

Solution:

In the first step we are going to apply a trigonometric formula for cos² x

cos² x = (1 + cos 2x)/2

= ∫ [ cos² 2 x- sin 6 x ] dx

= ∫ [ cos² 2 x] dx  - ∫ [ sin 6 x ] dx

= ∫ [(1 + cos 2x)/2] dx  - ∫ [ sin 6 x ] dx

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

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

Question 9

Integrate the following with respect to x,1/(1+sin x)

Solution:

Here we are going to multiply the given function by using the conjugate of the denominator

= ∫ [1/(1+sin x)] x (1-sin x)/(1-sinx) dx

= ∫ [(1-sin x)/(1+sin x)(1-sin x)] dx

= ∫ [(1-sin x)/(1²-sin²x)] dx

= ∫ [(1-sin x)/cos²x] dx

= ∫ [(1/cos²x) - (sin x/cos²x)] dx

= ∫ [sec²x - (sin x/cosx) (1/cos x)] dx

= ∫ [sec²x - tan x sec x] dx

= ∫ sec²x dx - ∫ tan x sec x dx

= tan x - sec x + C

Question 10

Integrate the following with respect to x,1/(1 - cos x)

Solution:

Here we are going to multiply the given function by using the conjugate of the denominator

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

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

= ∫ [(1+cos x)/(1²-cos²x)] dx

= ∫ [(1+cos x)/sin²x] dx

= ∫ [(1/sin²x) + (cos x/sin²x)] dx

= ∫ [cosec²x + (cos x/sin x) (1/sin x)] dx

= ∫ [cosec²x + cot x cosec x] dx

= ∫ cosec²x dx + ∫ cot x cosec x dx

= - cot x - cosec x + C

Question 11

Integrate the following with respect to x,√(1 - sin2x)

Solution:

now we are going to use a trigonometric formula for sin 2 x

sin 2x = 2 sin x cos x

= ∫ √(1 - sin2x) dx

= ∫ √(cos² x + sin²x - 2 sin x cos x) dx

= ∫ √[(cos x - sin x)²] dx

= ∫ (cos x - sin x) dx

= ∫ cos x dx - ∫ sin x dx

= sin x - (-cos x) + C

= sin x + cos x + C

integration worksheet4 solution3 integration worksheet4 solution3