Question

${\int }_0^{\pi }\frac{dx}{1+2{\sin }^{2}x}$

Answer 1

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

Let $\sin x=\frac{\tan x}{\sec x}$

We know, ${\sec }^{2}x={\tan }^{2}x+1$

Thus $2{\sin }^{2}x+1=2\frac{{\tan }^{2}x}{{\sec }^{2}x}+1=\frac{2{\tan }^{2}x+{\sec }^{2}x}{{\sec }^{2}x}=\frac{2{\tan }^{2}x+{\tan }^{2}x+1}{{\sec }^{2}x}=\frac{3{\tan }^{2}x+1}{{\sec }^{2}x}$

$\therefore I={\int }_0^{\pi }{\sec }^{2}x.\frac{1}{3{\tan }^{2}x+1}dx$

Substitute , $u=\tan x\to dx=\frac{du}{{\sec }^{2}x}$

Thus, $I={\int }_0^{\pi }\frac{1}{3u^2+1}du$

Substitute, $v=\sqrt{3}u\to du=\frac{du}{\sqrt{3}}$

$I=\frac{1}{\sqrt{3}}{\int }_0^{\pi }\frac{dv}{v^2+1}$

$I=\frac{1}{\sqrt{3}}ta{n}^{-1}v{|}_0^{\pi }$

$I=\frac{1}{\sqrt{3}}ta{n}^{-1}\sqrt{3}u{|}_0^{\pi }$

$I=\frac{1}{\sqrt{3}}ta{n}^{-1}\sqrt{3}\tan x{|}_0^{\pi }$

$I=\frac{1}{\sqrt{3}}ta{n}^{-1}\sqrt{3}\tan x{|}_0^{\pi }$

$I=\frac{1}{\sqrt{3}}ta{n}^{-1}\sqrt{3}\tan \pi -\frac{1}{\sqrt{3}}ta{n}^{-1}\sqrt{3}\tan 0$

$I=\frac{1}{\sqrt{3}}ta{n}^{-1}0-\frac{1}{\sqrt{3}}ta{n}^{-1}0$

$I=0$