Question

${\int }_0^{\frac{\pi }{2}}\frac{\sin {}^{2}x}{\sin x+\cos x}dx=-\frac{1}{2\sqrt{2}}\log (-2\sqrt{2}+3)$

Answer 1

${\int }_{\pi /2}^0\frac{{\sin }^{2}x}{\sin x+\cos x}dx$

${\int }_{\pi /2}^0\frac{(1-\cos 2x)/2}{\sin x+\cos x}dx$

$=\underset{I_1}{\frac{1}{2}\int \frac{1}{\sin x+\sin x}dx}-\underset{I_2}{\int \frac{\cos 2x}{\sin x+\cos x}dx}$

$I_1\Rightarrow \int \frac{1}{\sin x+\cos x}dx$ is $\tan \frac{x}{2}$

$\int \frac{2}{1+2\mu -{\mu }^2}d\mu$

$=2\int \frac{1}{-(\mu -1{)}^2+2}$

$=\frac{2}{\sqrt{2}}(\frac{In|(\mu -1)/\sqrt{2}-1|}{2}-\frac{In|(\mu -1)/\sqrt{2}|-1}{2})$

$=\sqrt{2}[\frac{1}{2In\left|(\frac{\tan x/2-1}{\sqrt{2}}+1)\right|}-\frac{1}{2}In\left|\tan \frac{x-1}{\sqrt{2}}\right|]$

$I_2\Rightarrow \int \frac{\cos 2x}{\sin x+\cos x}dx$

$=\int \frac{{\cos }^{2}x-{\sin }^{2}x}{\sin x+\cos x}dx=\int (\cos x-\sin x)dx$

$=\sin x+\cos x$

$\Rightarrow \frac{\sqrt{2}}{2}In\left|\frac{\tan x+1}{\sqrt{2}}\right|-In\left|\frac{\tan x/2-1}{\sqrt{2}}\right|-\sin x+\cos$

$I\Rightarrow \frac{-1}{2}-\frac{1}{2}\left(\frac{In(-\frac{1}{\sqrt{2}+1})-In\left(\frac{1}{\sqrt{2}+1}\right)}{\sqrt{2}}\right)-1$

$I=-\frac{In(3-2\sqrt{2})}{2\sqrt{2}}$

$=-\frac{1}{2\sqrt{2}}In(3-2\sqrt{2})$