Question

The value of ${\int }_0^{\pi }\frac{dx}{1+2{\sin }^{2}x}$ is

• A. $0$
• B. $\frac{\pi }{2\sqrt{3}}$
• C. $\frac{\pi }{\sqrt{3}}$
• D. none of these

Answer 1

Answer: (C)

$\frac{\pi }{\sqrt{3}}$

Let $I={\int }_0^{\pi }\frac{dx}{1+2{\sin }^{2}x}$
Since, $\sin (\pi -x)=\sin x$
$=2{\int }_0^{\frac{\pi }{2}}\frac{dx}{1+2{\sin }^{2}x}$
$=2{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}xdx}{{\sec }^{2}x+2{\tan }^{2}x}=2{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}xdx}{1+3{\tan }^{2}x}$
Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
So, $I=2{\int }_0^{\infty }\frac{dt}{1+3t^2}=\frac{2}{\sqrt{3}}[{\tan }^{-1}\left(\sqrt{3}t\right){]}_0^{\infty }$
$I=\frac{\pi }{\sqrt{3}}$