Question:

<div class='solution'> <p>Let $I=\int_0^{\pi/2}\frac{\sin^2 x}{\sin x+\cos x}dx$, $J=\int_0^{\pi/2}\frac{\cos^2 x}{\sin x+\cos x}dx$.</p> <p>By King, $I=J$. And $I+J=\int_0^{\pi/2}\frac{\sin^2x+\cos^2x}{\sin x+\cos x}dx=\int_0^{\pi/2}\frac{dx}{\sin x+\cos x}$.</p> <p>$\sin x+\cos x=\sqrt{2}\sin(x+\pi/4)$. $\int_0^{\pi/2}\frac{dx}{\sqrt{2}\sin(x+\pi/4)}=\frac{1}{\sqrt{2}}\int_{\pi/4}^{3\pi/4}\csc u\,du=\frac{1}{\sqrt{2}}[\ln|\csc u-\cot u|]_{\pi/4}^{3\pi/4}=\frac{1}{\sqrt{2}}\ln(\sqrt{2}+1)\cdot 2$.</p> <p>So $2I=\frac{\sqrt{2}\ln(\sqrt{2}+1)}{\sqrt{2}}\Rightarrow I=\frac{\ln(1+\sqrt{2})}{\sqrt{2}}$.</p> </div>