Evaluating $\int\limits_0^{\pi} \frac{dx}{1+2\sin^2x}$ 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 !
 A: Your integral is twice the below integral. The reason why your integral is twice the below integral is because it satisfies $f(2a-x)=f(x)$ where $a =\frac{\pi}{2}$. 


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*Kindly refer this link: How do i prove $\int_{0}^{2a}f(x) dx = \int_{0}^{a}f(x)dx +\int_{0}^{a}f(2a-x)dx$
\begin{align*}
\int_{0}^{\pi/2} \frac{1}{1+2\sin^{2}(x)} &= \int_{0}^{\pi/2} \frac{\sec^{2}(x)}{\sec^{2}(x)+2\tan^{2}(x)} 
\\&= \int_{0}^{\pi/2} \frac{\sec^{2}(x)}{1+3\tan^{2}(x)}
\end{align*}
A: I think that Weierstrass substitution could help. Let $$I=\int \dfrac{dx}{1+2\sin^2x}$$ So, using $t=\tan(\frac{x}{2})$, you arrive to $$I=\int \frac{2 \left(t^2+1\right)}{t^4+10 t^2+1}dt$$ Now, notice that the denominator write $$t^4+10 t^2+1=(t^2+\alpha)(t^2+\beta)$$ with $\alpha=2 \sqrt{6}+5$, $\beta=2 \sqrt{6}-5$. Use partial fraction decomposition and integrate; you should arrive to $$I=\frac{\tan ^{-1}\left(\left(\sqrt{3}-\sqrt{2}\right) t\right)+\tan
   ^{-1}\left(\left(\sqrt{3}+\sqrt{2}\right) t\right)}{\sqrt{3}}=\frac{\tan ^{-1}\left(\frac{2 \sqrt{3} t}{1-t^2}\right)}{\sqrt{3}}$$ Now, use the bounds for $t$.
A: We solve the problem in a way which is almost identical to yours, with a small correction. Let $F(x)$ be an antiderivative of our function. Then the integral is $F(\pi)-F(0)$.
The only problem with your solution is that for $\frac{\pi}{2}\lt x\lt \pi$, the function
$$\frac{\arctan(\tan(\sqrt{3}x))}{\sqrt{3}}\tag{1}$$
is not the correct antiderivative. The issue is the back substitution. 
For recall that $\arctan t$ is the number $t$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ whose tan is $t$. If we want an angle between $\frac{\pi}{2}$ and $\pi$, we need to use $\pi+\arctan t$. In particular, for $x$ near to but below $\pi$, $F(X)$ is given by
$$F(x) =\frac{\pi +\arctan(\tan(\sqrt{3}x))}{\sqrt{3}}.\tag{2}$$
Now "plug in" $\pi$ (in the right expression (2) and take away the result of plugging in $0$ (in the right expression (1)). We get $\frac{\pi}{\sqrt{3}}$. 
Remark: I would in fact break up the integral into two parts, as a reflexive response to the symmetry. And even if did not, for the substitution I would consider the two intervals separately, since singularities make me nervous.   However, that is not necessary. And in the solution above, $\frac{\pi}{2}$ does not even come into the picture. 
A: $\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}

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

A: You may write 
$$
\begin{align}
\int_0^{\pi/2}\dfrac{dx}{1+2\sin^2x}& =\int_0^{\pi/2}\dfrac{dx}{1+2(1-\cos^2x)}\\\\
&=\int_0^{\pi/2}\dfrac{dx}{3-2\cos^2x}\\\\
&=\int_0^{\pi/2}\dfrac{dx}{3-\frac{2}{1+\tan^2x}}\\\\
&=\int_0^{\pi/2}\dfrac{(1+\tan^2x)}{1+3\tan^2x}dx\\\\
&=\int_0^{\infty}\dfrac{du}{1+3u^2}\\\\
&=\left.\dfrac{\arctan(\sqrt{3}u)}{\sqrt{3}}\right|_{0}^{\infty}\\\\
&=\dfrac{\pi}{2\sqrt{3}}
\end{align}
$$ and use André Nicolas' comment: your initial integral is twice the preceding.
