#### Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 3

$\frac{\pi^{2}}{4}$

Hint:

$\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$

Given:

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

Explanation:

Let

$I=\int_{0}^{\pi} \frac{x \tan x}{\sec x+\cos x} d x, \quad\left[\therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$

$=\int_{0}^{\pi} \frac{-(\pi-x) \tan x}{-\sec x-\cos x} d x$

\begin{aligned} &I=\int_{0}^{\pi} \frac{(\pi-x) \tan x}{\sec x+\cos x} d x \\\\ &I=\int_{0}^{\pi} \frac{\pi \tan x}{\sec x+\cos x} d x-\int_{0}^{\pi} \frac{x \tan x}{\sec x+\cos x} d x \end{aligned}

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

Put

\begin{aligned} &\cos x=t \\\\ &-\sin x \; d x=d t \end{aligned}

When $x=0$ then $t=1$

When $x=\pi$ then $t=-1$

$=\pi \int_{1}^{-1} \frac{-d t}{1+t^{2}}$

$=\frac{\pi}{2} \int_{-1}^{1} \frac{d t}{1+t^{2}}, \quad\left[\therefore-\int_{b}^{a} f(x) d x=\int_{a}^{b} f(a-x) d x\right]$

\begin{aligned} &=\frac{\pi}{2}\left[\tan ^{-1} t\right]_{-1}^{1} \\\\ &=\frac{\pi}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(-1)\right] \\\\ &=\frac{\pi}{2}\left[\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right] \end{aligned}

\begin{aligned} &=\frac{\pi}{2}\left(\frac{\pi}{2}\right) \\\\ &=\frac{\pi^{2}}{4} \end{aligned}