Evaluating $\int\limits_0^{\pi} \frac{dx}{1+2\sin^2x}$

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.

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 $a =\frac{\pi}{2}$.

• 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*}

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 $t$.