# On the way of evaluate $\int_{0}^{\pi}\frac{dx}{1+2\sin^{2}x}$. [duplicate]

Evaluate $$\int_{0}^{\pi}\frac{dx}{1+2\sin^2x}$$

My approach $$\Longrightarrow\int_{0}^{\pi}\frac{dx}{1+2\sin^{2}x}=2\int_{0}^{\frac{\pi}{2}}\frac{\sec^{2}x}{1+3\tan^{2}x}dx=\frac{2}{\sqrt{3}}\left[\tan^{-1}\left(\sqrt{3}\tan x\right)\right]_{0}^{\frac{\pi}{2}}$$ $\tan x$ is undefined at $\dfrac{\pi}{2}$.

So I need to solve $\lim_{x\rightarrow\frac{\pi}{2}}$$\left[\tan^{-1}\left(\sqrt{3}\tan x\right)\right]$. I don't know how to solve.

## marked as duplicate by Jack, Nosrati, José Carlos Santos, Stefan4024, Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 2 '17 at 0:28

Note that as $x \to \frac{\pi}{2}$, $\sqrt{3}\tan(x) \to \infty$, therefore $\tan^{-1}(\infty) = \frac{\pi}{2}$