Get Answers to all your Questions

header-bg qa

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

Answers (1)

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}}

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads