#### Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 28 maths.

Answer: $\frac{\pi}{2ab}$

Hint: We use indefinite integral formula then put limits to solve this integral.

Given:$\int_{0}^{\frac{\pi}{2}} \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x$

Solution:

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

Dividing numerator and denominator by cos2x , we get

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\frac{1}{\cos ^{2} x}}{\frac{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x}{\cos ^{2} x}} d x \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\frac{a^{2} \sin ^{2} x}{\cos ^{2} x}+\frac{b^{2} \cos ^{2} x}{\cos ^{2} x}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{1}{\cos x}=\sec x\right] \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{a^{2} \tan ^{2} x+b^{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{\sin x}{\cos x}=\tan x\right] \end{aligned}

Put $\tan x=t$

$\Rightarrow sec^2xdx=dt$

When x=0  then t=0  and when $x=\frac{\pi}{2}$  then $t=\infty$

\begin{aligned} &I=\int_{0}^{\infty} \frac{1}{a^{2} t^{2}+b^{2}} d t \\ &=\frac{1}{a^{2}} \int_{0}^{\infty} \frac{1}{t^{2}+\frac{b^{2}}{a^{2}}} d t \\ &=\frac{1}{a^{2}} \int_{0}^{\infty} \frac{1}{t^{2}+\left(\frac{b}{a}\right)^{2}} d t \end{aligned}

\begin{aligned} &=\frac{1}{a^{2}}\left[\frac{1}{b} \tan ^{-1}\left(\frac{a t}{b}\right)\right]_{0}^{\infty} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)\right] \\ &=\frac{1}{a^{2}} \frac{a}{b}\left[\tan ^{-1}\left(\frac{a t}{b}\right)\right]_{0}^{\infty} \\ &=\frac{1}{a b}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right] \\ &=\frac{1}{a b}\left[\frac{\pi}{2}-0\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} \infty=\frac{\pi}{2}, \tan ^{-1} 0=0\right] \end{aligned}

$=\frac{1}{ab}\frac{\pi}{2}$

$=\frac{\pi}{2ab}$