Question

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

Answer 1

Given,

${\int }_0^{\frac{\pi }{2}}\frac{1}{a^2{\sin }^2\left(x\right)+b^2{\cos }^2\left(x\right)}dx$

substitute $x={\tan }^{-1}u$

$=\int \frac{1}{a^2{u}^2+b^2}du$

substitute $u=\frac{b}{a}v$

$=\int \frac{1}{ab(v^2+1)}dv$

$=\frac{1}{ab}{\tan }^{-1}v$

$=\frac{1}{ab}{\tan }^{-1}\left(\frac{a}{b}\tan \right(x\left)\right)+c$

$={\left[\frac{1}{ab}{\tan }^{-1}\left(\frac{a}{b}\tan \right(x\left)\right)\right]}_0^{\frac{\pi }{2}}$

$=\frac{\pi |a|}{2a^2|b|}-0$

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