Question

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

Answer 1

Using the property, ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

Applying this property to the given integral,

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

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

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

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

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

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

$\Rightarrow I=2\times \frac{\pi }{2}{\int }_0^{\pi }\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$ using the property,${\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f(a-x)dx$ if $f(2a-x)=f\left(x\right)$

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

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

$\Rightarrow I=\pi {\int }_0^{\frac{\pi }{2}}\frac{\frac{1}{{\cos }^{2}x}dx}{a^2\frac{{\cos }^{2}x}{{\cos }^{2}x}+b^2\frac{{\sin }^{2}x}{{\cos }^{2}x}}$

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

Put $b\tan x=t\Rightarrow b{\sec }^{2}xdx=dt$ or ${\sec }^{2}xdx=\frac{dt}{b}$

Also, when $x=0,t=0$ and when $x=\frac{\pi }{2},t\to \infty$

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

$\Rightarrow I=\frac{\pi }{ab}{\left|{\tan }^{-1}\left(\frac{t}{a}\right)\right]}_0^{\infty }$

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

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