Question

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

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{3\sqrt{3}}$
• C. $\frac{\pi }{\sqrt{3}}$
• D. $0$

Answer 1

Answer: (C)

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

${\int }_0^{\pi }\frac{dx}{1+2{\sin }^{2}x}$
$=2{\int }_0^{\pi /2}\frac{dx}{1+2{\sin }^{2}x}[\because {\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dxiff(2a-x)=f(x\left)\right]$

Dividing Numerator and denominator by $co{s}^{2}x$
$=2{\int }_0^{\pi /2}\frac{{\sec }^{2}x}{{\sec }^{2}x+2{\tan }^{2}x}dx=2{\int }_0^{\pi /2}\frac{{\sec }^{2}x}{1+3{\tan }^{2}x}dx$
put $\tan x=t$
$=2{\int }_0^{\infty }\frac{dt}{1+3t^2}$
$=2\times \frac{1}{\sqrt{3}}{\left[{\tan }^{-1}t\sqrt{3}\right]}_0^{\infty }$
$=\frac{2}{\sqrt{3}}\times \frac{\pi }{2}=\frac{\pi }{\sqrt{3}}$