#### Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 57

Answer:  $\frac{\pi}{\sqrt{35}}$

Hint: To solve this equation, we convert cos x into tan x

Given:  $\int_{0}^{\pi} \frac{d x}{6-\cos x}$

Solution

$I= \int _{0}^{\pi}\frac{dx}{6-\left ( \frac{1-\tan ^{2}\frac{x}{2}}{1+\tan\frac{x}{2}} \right )}$

$=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{6+6 \tan ^{2} \frac{x}{2}-1+\tan ^{2} \frac{x}{2}} d x$

\begin{aligned} &=\frac{1}{7} \int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{\frac{5}{7}+\tan ^{2} \frac{x}{2}} d x \\ & \end{aligned}

$\text { Let } \tan \frac{x}{2}=t \\$

$\frac{1}{2} \sec ^{2} \frac{x}{2} d x=d t$

\begin{aligned} &\sec ^{2} \frac{x}{2} d x=2 d t \\ & \end{aligned}

$I=\frac{2}{7} \int_{0}^{\pi} \frac{d t}{\sqrt{\frac{5}{7}}^{2}+t^{2}}$

\begin{aligned} &I=\frac{2}{7} \frac{1}{\sqrt{\frac{5}{7}}}\left(\tan ^{-1} \frac{t}{\sqrt{\frac{5}{7}}}\right)_{0}^{\infty} \\ & \end{aligned}

$=\frac{2}{\sqrt{35}}\left(\frac{\pi}{2}-0\right) \\$

$I=\frac{\pi}{\sqrt{35}}$