#### Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 36

$\frac{\pi }{4}$

Hint:

To solve this equation we convert cos in term of tan.

Given:

$\int_{0}^{\pi} \frac{1}{5+3 \cos x} d x$

Solution:

Let

\begin{aligned} &I=\int_{0}^{\pi} \frac{1}{5+3 \cos x} d x \\\\ &\therefore \cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} \end{aligned}

$\cos \theta=\frac{1-\tan ^{2} \theta / 2}{1+\tan ^{2} \theta / 2}$

$I=\int_{0}^{\pi} \frac{1}{5+3\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$

\begin{aligned} &I=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{8+2 \tan ^{2} \frac{x}{2}} d x \\\\ &\text { Let } \tan \frac{x}{2}=t, \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t \end{aligned}

\begin{aligned} &=\frac{2}{2} \int_{0}^{\infty} \frac{d t}{4+t^{2}} \\\\ &=\int_{0}^{\infty} \frac{1}{(2)^{2}+t^{2}} d t \end{aligned}

\begin{aligned} &=\left[\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)\right]_{0}^{\infty} \\\\ &=\frac{1}{2}\left(\frac{\pi}{2}\right)=\frac{\pi}{4} \end{aligned}