Question

Evaluate: ${\int }_0^{\pi /2}\frac{x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$.

Answer 1

${\int }_0^{\pi /2}\frac{x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$

Divide numinator and denominator by $co{s}^{2}x$

${\int }_0^{\pi /2}\frac{x{\sec }^{2}x}{a^2+b^2{\tan }^{2}x}dx$.

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

$=\frac{\pi }{2ab}[{\tan }^{-1}\frac{b}{a}{\tan }^{-1}\pi -{\tan }^{-1}\frac{b}{a}\tan 0)$

$=\frac{\pi }{2ab}[{\tan }^{-1}0-{\tan }^{-1}0]$

$=0$.