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

Answer:  $\frac{\pi^{2}}{2ab}$

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

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

Solution:

\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}

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

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