Question

Prove that ;
${\int }_0^{\pi }\frac{xdx}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}=\frac{{\pi }^2}{2ab}$

Answer 1

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

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

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

Adding $\left(1\right)$ and $\left(2\right)$ we get

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

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

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

Divide the numerator and denominator by ${\cos }^{2}x$ we get

$\Rightarrow I=\frac{\pi }{2}{\int }_0^{\pi }\frac{{\sec }^{2}xdx}{a^2+b^2{\tan }^{2}x}$

$\Rightarrow I=\pi {\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}xdx}{a^2+b^2{\tan }^{2}x}$ using ${\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$

Let $b\tan x=t\Rightarrow b{\sec }^{2}xdx=dt$

When $x=0\Rightarrow t=0$ and when $x=\frac{\pi }{2}\Rightarrow t=\infty$

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

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

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

$=\frac{\pi }{ab}\frac{\pi }{2}=\frac{{\pi }^2}{2ab}$