Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 11

Answers (1)

best_answer

Answer: $2-\frac{\pi}{2}$

Given: $\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{(1+\cos x)^{2}} d x$

Hint:

Use the formula and then apply integration rule by parts method.

Solution:

$\begin{aligned} &\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{(1+\cos x)^{2}} d x \\ & \end{aligned}$

$=\int_{0}^{\frac{\pi}{2}} \frac{1-\cos ^{2} x}{(1+\cos x)^{2}} d x$

$\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{1-\cos x}{(1+\cos x)} d x \\ & \end{aligned}$

$=\int_{0}^{\frac{\pi}{2}} \frac{1-\cos x-1+1}{1+\cos x} d x$

$\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{2-(1+\cos x)}{1+\cos x} d x \\ & \end{aligned}$

$=\int_{0}^{\frac{\pi}{2}} \frac{2}{1+\cos x} d x-\int_{0}^{\frac{\pi}{2}} \frac{1+\cos x}{1+\cos x} d x$

$\begin{aligned} &=2 \int_{0}^{\frac{\pi}{2}} \frac{(1-\cos x)}{\left(1-\cos ^{2} x\right)} d x-\frac{\pi}{2} \\ & \end{aligned}$

$=2 \int_{0}^{\frac{\pi}{2}} \frac{(1-\cos x)}{\sin ^{2} x} d x-\frac{\pi}{2}$

$\begin{aligned} &=2 \int_{0}^{\frac{\pi}{2}} \cos e c^{2} x-\operatorname{cosec} x \cdot \cot x d x-\frac{\pi}{2} \\ & \end{aligned}$

$=2(-\cot x+\cos e c x)_{0}^{\frac{\pi}{2}}-\frac{\pi}{2} \\$

$=2(-0+1+0)-\frac{\pi}{2} \\$

$=2-\frac{\pi}{2}$

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

‘Feeling Devastated’: CBSE 12th students allege low marks due to on-screen marking; call for manual evaluation

anam-khan

'No separate prep for CBSE and JEE': Gurugram CBSE topper balances both with one routine, here's how

anam-khan

What does RT mean in CBSE result?

Latest Question