TITLE: Integrating $\frac{x dx}{\sin x+\cos x}$
QUESTION [8 upvotes]: I am trying to carry out this integration but I seem to be going wrong:
$$I=\int_{0}^{\frac{\pi}{2}}\frac{x dx}{\sin x+\cos x}=\int_{0}^{\frac{\pi}{2}}\frac{(\frac{\pi}{2}-x) dx}{\sin(\frac{\pi}{2}-x)+\cos (\frac{\pi}{2}-x)} \implies 2I=\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{ dx}{\sin x+\cos x}$$ 
I am not able to proceed from here.

REPLY [4 votes]: $$\dfrac1{\sin(x) + \cos(x)} = \dfrac1{\sqrt{2}} \left(\dfrac1{\dfrac1{\sqrt{2}}\sin(x) + \dfrac1{\sqrt{2}}\cos(x)} \right) = \dfrac1{\sqrt{2}} \sec(x - \pi/4)$$ Now integrate, $$\int_0^{\pi/2} \dfrac{dx}{\sin(x) + \cos(x)} = \dfrac1{\sqrt{2}}\int_0^{\pi/2} \sec(x-\pi/4)dx = \dfrac1{\sqrt{2}}\int_{-\pi/4}^{\pi/4} \sec(x)dx$$ and finish it off.