Question

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{a^2.{\sin }^{2}x+b^2.{\cos }^{2}x}dx$ is equal to

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

Answer 1

The correct option is D $\frac{\pi }{2ab}$

$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{a^2.{\sin }^{2}x+b^2.{\cos }^{2}x}dx$

$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sec }^{2}x}{a^2.{\tan }^{2}x+b^2}dx$

Let $\tan x=t,{\sec }^{2}xdx=dt$

$x\to 0\Rightarrow t\to 0,x\to \frac{\pi }{2}\Rightarrow t\to \infty$

$I=\underset{0}{\overset{\infty }{\int }}\frac{dt}{b^2+a^2{t}^2}$

$=\frac{1}{ab}[{\tan }^{-1}\frac{at}{b}{]}_0^{\infty }$

$=\frac{1}{ab}(\frac{\pi }{2}-0)$

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