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Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 50

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Answer:  \frac{\pi^{2}}{2 a \sqrt{a^{2}-1}}

Hint: To solve this equation we convert cos

Given:  \int_{0}^{\pi} \frac{x}{a^{2}-\cos ^{2} x} d x, a>1


Let   I=\int_{0}^{\pi} \frac{x}{a^{2}-\cos ^{2} x} d x

\begin{aligned} &I=\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\ & \end{aligned}        

I=\int_{0}^{\pi} \frac{\pi-x}{a^{2}-\cos ^{2}(\pi-x)} d x=\int_{0}^{\pi} \frac{\pi-x}{a^{2}-\cos ^{2} x} d x

\begin{aligned} &2 I=\int_{0}^{\pi} \frac{\pi}{a^{2}-\cos ^{2} x} d x \\ & \end{aligned}

2 I=\pi \int_{0}^{\pi} \frac{1}{2 a}\left[\frac{1}{1+a \cos x}+\frac{1}{1-\cos x}\right] d x \\

2 I=\frac{\pi}{2 a} \int_{0}^{\pi} \frac{1}{a+\cos x} d x+\frac{\pi}{2 a} \int_{0}^{\pi} \frac{1}{1-a \cos x} d x

\begin{aligned} &2 I=\frac{\pi}{a} \int_{0}^{\pi} \frac{1}{a+\cos x} d x \\ & \end{aligned}

\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}

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

2 I=\frac{\pi}{a} \int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{(a-1) \tan ^{2} \frac{x}{2}+a+1} d x

\begin{aligned} &2 I=\frac{\pi}{a} \int_{0}^{\pi} \frac{\frac{d}{d x} \tan \frac{x}{2} d x}{(a-1) \tan ^{2} \frac{x}{2}+(a+1)} \\ & \end{aligned}

I=\frac{\pi}{a(a-1)} \int_{0}^{\pi} \frac{d \tan \frac{x}{2}}{\tan ^{2} \frac{x}{2}+\frac{a+1}{a-1}}

\begin{aligned} &I=\frac{\pi}{a(a-1)} \times \sqrt{\frac{a+1}{a+1}}\left|\tan ^{-1} \frac{\tan \frac{x}{2}}{\sqrt{\frac{a+1}{a-1}}}\right|_{0}^{\pi} \\ & \end{aligned}

I=\frac{\pi}{a \sqrt{a^{2}-1}}\left(\tan ^{-1} \infty-\tan ^{-1} 0\right)

\begin{aligned} &=\frac{\pi}{a} \times \frac{1}{a \sqrt{a^{2}-1}} \times \frac{\pi}{2} \\ & \end{aligned}

=\frac{\pi^{2}}{2 a \sqrt{a^{2}-1}}



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