Question

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

Answer 1

Let $I={\int }_0^{\pi }\frac{x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$
$\Rightarrow I={\int }_0^{\pi }\frac{\pi -x}{a^2{\cos }^2(\pi -x)+b^2{\sin }^2(\pi -x)}dx$
$[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]$
$\Rightarrow I={\int }_0^{\pi }\frac{\pi -x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$
$\Rightarrow I={\int }_0^{\pi }\frac{\pi }{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx-{\int }_0^{\pi }\frac{x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$
$\Rightarrow I=\pi {\int }_0^{\pi }\frac{1}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx-I$
$\Rightarrow 2I=\pi {\int }_0^{\pi }\frac{\pi }{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$
$\Rightarrow 2I=2\pi {\int }_0^{\frac{\pi }{2}}\frac{1}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$
$[\because {\cos }^2(\pi -x)={\cos }^{2}x,{\sin }^2(\pi -x)={\sin }^{2}x]$
$\Rightarrow I=\frac{\pi }{b^2}{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}x}{\frac{a^2}{b^2}+{\tan }^{2}x}$
Let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
Changing the limit of integral when $x=0\Rightarrow t=0$ and $x=\frac{\pi }{2}\Rightarrow t\to \infty$
Therefore,
$I=\frac{\pi }{b^2}{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}x}{\frac{a^2}{b^2}+{\tan }^{2}x}dx$
$=\frac{\pi }{b^2}{\int }_0^{\infty }\frac{dt}{{\left(\frac{a}{b}\right)}^2+t^2}$
$=\frac{\pi }{b^2}{\left[\frac{1}{\frac{a}{b}}{\tan }^{-1}\left(\frac{t}{\frac{a}{b}}\right)\right]}_0^{\infty }$
$=\frac{\pi }{ab}{\left[{\tan }^{-1}\left(\frac{bt}{a}\right)\right]}_0^{\infty }$
$=\frac{\pi }{ab}\left[{lim}_{t\to \infty }{\tan }^{-1}(\frac{bt}{a}-0)\right]$
$=\frac{\pi }{ab}\left[\frac{\pi }{2}\right]$
$\therefore I=\frac{{\pi }^2}{2ab}$
Hence, ${\int }_0^{\pi }\frac{x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx=\frac{{\pi }^2}{2ab}$