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This question already has an answer here:

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.

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marked as duplicate by Jack, Nosrati, José Carlos Santos, Stefan4024, Namaste calculus Dec 2 '17 at 0:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

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