## Integration Worksheet4 solution6

In this page integration worksheet4 solution6 we are going to see solution of some practice question from the worksheet of integration.

Question 21

Integrate the following with respect to x, e^(x log a) e^x

Solution:

= ∫e^(x log a) e^x dx

= ∫e^(log a^x) e^x dx

= ∫(a^x e^x) dx

= ∫(ae)^x dx

= [(ae)^x/log ae] + C

= [(ae)^x/[log a + log e] + C

= [(ae)^x/[1 + log a] + C

Question 22

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

Solution:

In the first step we are going to multiply and divide it by the conjugate of the denominator.

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

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

= ∫[(1 - sin x)²/(1²-sin² x)] dx

= ∫[(1 - sin x)²/cos² x dx

= ∫[(1 - sin x)/cos x]² dx

= ∫[(1/cos x) - (sin x/cos x)]² dx

= ∫[sec x - tan x]² dx

= ∫[sec² x + tan² x - 2 sec x tan x] dx

= ∫[sec² x + sec²x - 1 - 2 sec x tan x] dx

= ∫[2 sec² x - 2 sec x tan x - 1] dx

= ∫2 sec² x dx- 2 ∫sec x tan x dx  - ∫1 dx

= 2 tan x - 2 sec x  - x + C

Question 23

Integrate the following with respect to x, ) [√x + (1/√x)]²

Solution:

In the first step we are going to multiply and divide it by the conjugate of the denominator.

= ∫[√x + (1/√x)]²dx

= ∫[(√x)² + (1 /√x)² + 2 √x (1/√x)]  dx

= ∫[x + 1 /x + 2]  dx

= ∫ x dx  +∫ (1 /x) dx + ∫ 2 dx

= x²/2 + log x + 2 x + C

Question 24

Integrate the following with respect to x, sin m x cos n x

Solution:

We are going to multiply and divide the given question by 2

= ∫ (1/2) 2 sin m x cos n x dx

now we are going to us the formula for sin A cos B that is sin(A+B)-sin(A-B)

= ∫ (1/2) [ sin (m x + n x) - sin (m x - n x) ]dx

= ∫ (1/2) [ sin (m + n)x  - sin (m - n)x ]dx

= (1/2) ∫[ sin (m + n)x dx  - ∫sin (m - n)x dx

= (1/2) [-cos(m+n)x/(m+n)+cos(m-n)x/(m-n)] + C

integration worksheet4 solution6 integration worksheet4 solution6