Question

Show that $I={\int }_0^{\pi }\frac{xdx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}=\frac{\pi }{2ab}$

Answer 1

Let $I={\int }_0^{\pi }\frac{xdx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$ ...(1)

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$\therefore I={\int }_0^{\pi }\frac{(\pi -x)}{a^2{\cos }^2(\pi -x)+b^2{\sin }^2(\pi -x)}$

$\Rightarrow I={\int }_0^{\pi }\frac{(\pi -x)dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$ ...(2)

Adding (1) and (2), we get
$2I={\int }_0^{\pi }\frac{(x+\pi -x)dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}=\pi {\int }_0^{\pi }\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$

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

Divide numerator and denominator by ${\cos }^{2}x.$
$\therefore I=\pi {\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{a^2+b^2{\tan }^{2}x}$

Put $b\tan x=t$

$\therefore b{\sec }^{2}xdx=dt.$

When $x=0\Rightarrow t=0$

When $x=\frac{\pi }{2}\Rightarrow t=\infty$

$\therefore I=\frac{\pi }{b}{\int }_0^{\infty }\frac{dt}{a^2+t^2}$

$=\frac{\pi }{b}.\frac{1}{a}{\left[{\tan }^{-1}\frac{t}{a}\right]}_0^{\infty }$

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