Question

Solve: $I=2{\int }_0^{\pi /2}\frac{{\sec }^{2}x}{1+3{\tan }^{2}x}dx$

Answer 1

$I=2{\int }_0^{\pi /2}\frac{{\text{sec}}^{2}x}{1+3{\tan }^{2}x}dx=\frac{2}{3}{\int }_0^{\pi /2}\frac{{\text{sec}}^{2}x}{\frac{1}{3}+{\tan }^{2}x}dx=\frac{2}{3}[{\text{tan}}^{-1}(\sqrt{3}\tan x){]}_0^{\pi /2}=\frac{2\sqrt{3}}{3}[\frac{\pi }{2}-0]=\frac{\pi }{\sqrt{3}}$