TITLE: Evaluating a definite integral: $\int_0^{\pi}\frac{x}{1-\sin{x}\cos{x}}\,\mathrm dx$
QUESTION [8 upvotes]: Ok so I'm supposed to evaluate:
$$\int_0^{\pi}\frac{x}{1-\sin{x}\cos{x}}\,\mathrm dx$$
I tried using definite integral properties, but this didn't yield any good follow up, because I couldn't evaluate the resulting integrals. 
How exactly can one go about solving this? (preferably without contour integration)

REPLY [2 votes]: $$I=\int_0^\pi\frac{x}{1-\sin(x)\cos(x)}\ dx=\int_0^{\pi}\frac{2x}{2-\sin(2x)}\ dx$$
$$=\frac12\int_0^{2\pi}\frac{x}{2-\sin(x)}\ dx\overset{x+\pi/2=t}{=}\frac12\int_{\pi/2}^{5\pi/2}\frac{t-\pi/2}{2+\cos(t)}\ dt$$
Using the identity
$$\frac{1}{a+b\cos(t)}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nt)},\  a>b$$
We have 
$$\frac{1}{2+\cos(x)}=\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}\sum_{n=1}^{\infty}\left(\sqrt{3}-2\right)^n\cos{(nx)}$$
$$\Longrightarrow I=\frac{1}{2\sqrt{3}}\underbrace{\int_{\pi/2}^{5\pi/2}\left(t-\frac{\pi}{2}\right)\ dt}_{2\pi^2}+\frac{1}{\sqrt{3}}\sum_{n=1}^\infty (\sqrt{3}-2)^n\underbrace{\int_{\pi/2}^{5\pi/2}\left(t-\frac{\pi}2\right)\cos(nt)\ dt}_{A}$$
$$A=\int_{\pi/2}^{5\pi/2}t\cos(nt)\ dt-\frac{\pi}{2}\underbrace{\int_{\pi/2}^{5\pi/2}\cos(nt)\ dt}_{0}=\frac{2\pi}{n}\sin\left(\frac{\pi}{2}n\right), \quad n=1,2,3,...$$
$$\Longrightarrow I=\frac{\pi^2}{\sqrt{3}}+\frac{2\pi}{\sqrt{3}}\underbrace{\sum_{n=1}^\infty \frac{(\sqrt{3}-2)^n}{n}\sin\left(\frac{\pi}{2}n\right)}_{-\pi^2/12}=\frac{5\pi^2}{6\sqrt{3}}$$
Details for the last result;
$$\sum_{n=1}^\infty \frac{(\sqrt{3}-2)^n}{n}\sin\left(\frac{\pi}{2}n\right)=\Im\sum_{n=1}^\infty \frac{(\sqrt{3}-2)^ne^{in\pi/2}}{n}=\Im\sum_{n=1}^\infty \frac{[i(\sqrt{3}-2)]^n}{n}$$
$$=-\Im\ln\left(1-i(\sqrt{3}-2)\right)=-\tan^{-1}(-\sqrt{3}+2)=-\frac{\pi^2}{12}$$