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

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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:

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

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

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

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

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

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

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

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

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

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

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

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

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