Question

${\int }_0^{\pi /2}\frac{1}{a^2.si{n}^{2}x+b^2.co{s}^{2}x}dx$ is equal to

• A. $\frac{\pi a}{4b}$
• B. $\frac{\pi b}{4a}$
• C. $\frac{\pi a}{2b}$
• D. $\frac{\pi }{2ab}$

Answer 1

Answer: (D)

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

$I={\int }_0^{\pi /2}\frac{1}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$
$I={\int }_0^{\pi /2}\frac{1}{a^2{\cos }^{2}x[1+\frac{b^2}{a^2}{\tan }^{2}x]}dx$
$I={\int }_0^{\pi /2}\frac{1}{a^2[1+\frac{b^2}{a^2}{\tan }^{2}x]}dx$
Put, $\frac{b}{a}\cdot \tan x=u$
$\therefore \frac{b}{a}{\sec }^{2}x=\frac{du}{dx}$
$du=\frac{b}{a}{\sec }^{2}xdx\Rightarrow {\sec }^{2}xdx=\frac{a}{b}du$
Also, $x=0\Rightarrow u=0$ and $x=\frac{\pi }{2}\Rightarrow u=\infty$
$I={\int }_0^{\infty }\frac{a}{b}\cdot \frac{1.du}{a^2[1+u^2]}$
$I={\int }_0^{\infty }\frac{1}{ab}\cdot \frac{du}{1+u^2}$
$I=\frac{1}{ab}{\left|{\tan }^{-1}u\right|}_0^{\infty }=\frac{1}{ab}({\tan }^{-1}\infty -{\tan }^{-1}0)$
$I=\frac{1}{ab}[\frac{\pi }{2}-0]$
$I=\frac{\pi }{2ab}$