TITLE: Evaluating $\int\limits_0^{\pi} \frac{dx}{1+2\sin^2x}$
QUESTION [12 upvotes]: After making u substitution two times, I am getting indefinite integral as $$\int\limits\dfrac{dx}{1+2\sin^2x} = \dfrac{\arctan(\sqrt{3}\tan(x))}{\sqrt{3}}+ C$$
I am stuck at working the bounds because I am getting $0$ but the answer needs to be $\pi/\sqrt{3}$.
 I think $\arctan(x)$ is defined in $-\pi/2 \lt x\lt\pi/2$ while making u substitution. However I don't see how to use this observation to my advantage. Can anybody help in getting the answer and understanding whats going on. Thanks !

REPLY [3 votes]: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\color{#66f}{\large\int{\dd x \over 1 + 2\sin^{2}\pars{x}}}&
=\int{\sec^{2}\pars{x}\,\dd x \over \sec^{2}\pars{x} + 2\tan^{2}\pars{x}}
= \overbrace{\int{\sec^{2}\pars{x}\,\dd x \over 3\tan^{2}\pars{x} + 1}}
^{\color{#c00000}{\ds{\mbox{Set}\ t \equiv \tan\pars{x}}}}
\\[5mm]&={1 \over \root{3}} \ \underbrace{%
\int{\root{3}\,\dd t \over \pars{\root{3}t}^{2} + 1}}
_{\color{#c00000}{\ds{\root{3}t \equiv \xi\ \imp\ t = {\root{3} \over 3}\,\xi}}}\ =\
{\root{3} \over 3}\int{\dd\xi \over \xi^{2} + 1}
\\[5mm]&={\root{3} \over 3}\,\arctan\pars{\xi}={\root{3} \over 3}\,\arctan\pars{\root{3}t}
\end{align}

$$
\color{#66f}{\large\int{\dd x \over 1 + 2\sin^{2}\pars{x}}
={\root{3} \over 3}\,\arctan\pars{\root{3}\tan\pars{x}}} + \mbox{a constant}
$$