#### Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 51

Answer:   $I=\pi\left(\frac{\pi}{2}-1\right)$

Hint: To solve this equation we split tan x and sec x

Given:   $\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x$

Solution:  $I=\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x$

$I=\int_{0}^{\pi} \frac{x \frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}} d x$                                              $\left[\begin{array}{l} \because \tan x=\frac{\sin x}{\cos x} \\ \sec =\frac{1}{\cos x} \end{array}\right]$

$I=\int_{0}^{\pi} \frac{x \sin x}{1+\sin x} d x$

\begin{aligned} &I=\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\ & \end{aligned}

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

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

\begin{aligned} &I=\int_{0}^{\pi} \frac{\pi \sin x-x \sin x}{1+\sin x} d x \\ & \end{aligned}

$I=\int_{0}^{\pi} \frac{\sin x}{1+\sin x} d x-\int_{0}^{\pi} \frac{x \sin x}{1+\sin x} d x \\$

$I=\int_{0}^{\pi} \frac{\sin x}{1+\sin x} d x-I$

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

\begin{aligned} &2 I=\pi \int_{0}^{\pi} \frac{\sin x-\sin ^{2} x}{\cos ^{2} x} d x \\ & \end{aligned}

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

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

$I=\pi \int_{0}^{\pi} \sec x \tan x d x-\pi \int_{0}^{\pi} \tan ^{2} x d x$

\begin{aligned} &I=\pi \int_{0}^{\pi} \sec x \tan x d x-\pi \int_{0}^{\pi} \sec ^{2} x d x+\pi \int_{0}^{\pi} 1 d x \\ & \end{aligned}

$2 I=\pi[\sec x]_{0}^{\pi}-\pi[\tan x]_{0}^{\pi}+\pi[x]_{0}^{\pi}$

\begin{aligned} &2 I=\pi(\sec \pi-\sec 0)-\pi(\tan \pi-\tan 0)+\pi(\pi-0) \\ & \end{aligned}

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

$I=\frac{-2 \pi+\pi^{2}}{2} \\$

$I=\pi\left(\frac{-2}{2}+\frac{\pi}{2}\right) \\$

$I=\pi\left(\frac{\pi}{2}-1\right)$