Question

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

Answer 1

We have, $I={\int }_0^{\pi }\frac{x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$ .... (i)
$I={\int }_0^{\pi }\frac{\pi -x}{a^2{\cos }^2(\pi -x)+b^2{\sin }^2(\pi -x)}dx$
$(\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx)$
$I={\int }_0^{\pi }\frac{\pi -x}{a^2{\cos }^{2}x+b^2\sin x}dx$ .... (ii)
On adding equations (i) and (ii), we get
$2I={\int }_0^{\pi }\frac{x+\pi -x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$
$2I=\pi {\int }_0^{\pi }\frac{1}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$
Using the property, ${\int }_0^{2a}f\left(x\right)\cdot d={\int }_0^a\left[f\right(x)+f(2a-x\left)\right]dx$
$\therefore 2I=\pi \left[{\int }_0^{\pi /2}\frac{1}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx+{\int }_0^{\pi /2}\frac{1}{a^2{\cos }^2(\pi -x)+b^2{\sin }^2(\pi -x)}dx\right]$
$I=\pi {\int }_0^{\pi /2}\frac{1}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx=\pi {\int }_0^{\pi /2}\frac{dx}{a^2{\cos }^{2}x(1+\frac{b^2}{a^2}{\tan }^{2}x)}$
$=\frac{\pi }{a^2}{\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{1+\frac{b^2}{a^2}{\tan }^{2}x}$ [By putting $u=\frac{b}{a}\tan x,du=\frac{b}{a}{\sec }^{2}xdx$ when $x=0,u=0$ when $x=\frac{\pi }{2},u=\infty ]$
$=\frac{\pi }{a^2}{\int }_0^{\infty }\frac{\frac{a}{b}du}{1+u^2}$
$=\frac{\pi }{ab}{\int }_0^{\infty }\frac{du}{1+u^2}=\frac{\pi }{ab}\left[{\tan }^{-1}\right(u){]}_0^{\infty }+c$
$=\frac{\pi }{ab}[{\tan }^{-1}\infty -{\tan }^{-1}0]+c$
$=\frac{\pi }{ab}[\frac{\pi }{2}-0]+c$
$=\frac{{\pi }^2}{2ab}$