# Get Answers to all your Questions

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

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}

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

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

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

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support