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

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

Solution:

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}}

 

 

Posted by

infoexpert27

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