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Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 43 Maths Textbook Solution.

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Answer:  \frac{\pi^{2}}{2ab}

Hint: To solve this equation we will convert statement in terms of tan and sec.

Given:  \int_{0}^{\pi} \frac{x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} d x


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

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

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

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

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

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

I=\frac{\pi}{2} \int_{0}^{\pi} \frac{\sec ^{2} x}{a^{2}+b^{2} \tan ^{2} x} d x

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

I=\frac{\pi}{b} \int_{0}^{\infty} \frac{1}{a^{2}+t^{2} d t}                              \left[\begin{array}{l} b \tan x=t \\ b \sec ^{2} d x=d t \end{array}\right]                   

  I=\frac{\pi}{b}\left[\frac{1}{a} \tan ^{-} \frac{t}{a}\right]_{0}^{\infty}                                                 

\begin{aligned} &I=\frac{\pi}{a b}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right] \\ & \end{aligned}

I=\frac{\pi}{a b}\left[\frac{\pi}{2}-0\right] \\

I=\frac{\pi^{2}}{2 a b}


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