#### Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 2

2

Hint:

Using $\int x \; d x \text { and } \frac{d}{d x}(\tan x)$

Given:

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

Explanation:

Let

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

\begin{aligned} &=\int_{0}^{\pi} \frac{1+\tan ^{2 } \frac{x}{2}}{1+\tan ^{2} x / 2+2 \tan x / 2} d x \\\\ &=\int \frac{\sec ^{2 } \frac {x}{2}}{\tan ^{2} x / 2+2 \tan \frac{x}{2}+1} d x \end{aligned}

\begin{aligned} &\text { Put } 2 \tan \frac{x}{2}=t \\\\ &\sec ^{2} x / 2 \cdot 1 / 2 d x=d t \\\\ &\sec ^{2} x / 2 d x=2 d t \end{aligned}

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

When $x=\pi$ then $t= \infty$

\begin{aligned} &=\int_{0}^{\infty} \frac{2 d t}{t^{2}+2 t+1} \\\\ &=\int_{0}^{\infty} \frac{1}{(t+1)^{2}} d t \end{aligned}

\begin{aligned} &=\int_{0}^{\infty}(t+1)^{-2} d t \\\\ &=\left[\frac{(t+1)^{2}}{-2+1}\right]_{9}^{\infty} \end{aligned}

\begin{aligned} &=-2\left[\frac{1}{(t+1)}\right]_{9}^{\infty} \\\\ &=-2(0-1) \\\\ &=2 \end{aligned}