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

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}}

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