Compute $\int_{0}^{\frac{\pi}{2}}\frac{\sin^2x\cos x}{\sin x+\cos x}dx$ I'd like your help with computing the following integral:$$ \int_{0}^{\frac{\pi}{2}}\frac{\sin^2x\cos x}{\sin x+\cos x}dx.$$ I used $t=\sin x$ and got that $$\int_{0}^{\frac{\pi}{2}}\frac{\sin^2x\cos x}{\sin x+\cos x}dx=\int_{0}^{1}\frac{t^2}{t+\sqrt{1-t^2}}dt,$$  but I can't see how I it use either. I tried also to use $t=2\arctan x$, without any successes. 
What do you think?
Thank you very much!

The light: Denote $\displaystyle I = \int^{\pi/2}_0 \frac{ \sin^2 x \cos x}{\sin x + \cos x} dx .$ Let $ u= \pi/2 -x $ to find $\displaystyle  I = \int^{\pi/2}_0 \frac{\cos^2 x \sin x}{\sin x+\cos x} dx .$ Adding these two gives $\displaystyle  2I = \int^{\pi/2}_0 \frac{ \sin x \cos x (\sin x+ \cos x) }{\sin x + \cos x} dx = \int^{\pi/2}_0 \sin x \cos x dx = \frac{\sin^2 x}{2} \biggr|^{\pi/2}_0 = \frac{1}{2}$ so $ I = \frac{1}{4}.$

The dark path: Applying the identity $ \sin x + \cos x = \sqrt{2} \sin ( x +\pi/4) $ gives $\displaystyle  I = \frac{1}{\sqrt{2}} \int^{\pi/2}_0 \frac{\sin^2 x \cos x}{\sin (x+\pi/4)} dx = \frac{1}{\sqrt{2}} \int^{3\pi/4}_{\pi/4} \frac{\sin^2 (u-\pi/4) \cos (u-\pi/4)}{\sin u} du.$ Now apply $\displaystyle \sin^2 t = \frac{1-\cos 2t}{2}$  and the addition theorems, the result will eventually come.
