Question

Evaluate : ${\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x}{\sin x+\cos x}dx$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x}{\sin x+\cos x}dx$ ... (i)

$\therefore I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^2(\frac{\pi }{2}-x)}{\sin (\frac{\pi }{2}-x)+\cos (\frac{\pi }{2}-x)}dx$

$[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{\cos x+\sin x}dx$ ... (ii)

Adding (i) and (ii), we get

$2I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x+\cos x}dx$

$={\int }_0^{\frac{\pi }{2}}\frac{1}{\sin x+\cos x}dx$

$=\frac{1}{\sqrt{2}}{\int }_0^{\pi /2}\frac{dx}{\cos x.\frac{1}{\sqrt{2}}+\sin x.\frac{1}{\sqrt{2}}}$

$=\frac{1}{\sqrt{2}}{\int }_0^{\pi /2}\frac{dx}{\cos (x-\frac{\pi }{4})}$

$=\frac{1}{\sqrt{2}}{\int }_0^{\pi /2}\sec (x-\frac{\pi }{4})dx$

$=\frac{1}{\sqrt{2}}{\left[\log |\sec (x-\frac{\pi }{4})+\tan (x-\frac{\pi }{4})|\right]}_0^{\pi /2}$

$=\frac{1}{\sqrt{2}}[\log |\sec \frac{\pi }{4}+\tan \frac{\pi }{4}|-\log |\sec \left(\frac{-\pi }{4}\right)+\tan \left(\frac{-\pi }{4}\right)|]$

$=\frac{1}{\sqrt{2}}[\log |\sqrt{2}+1|-\log |\sqrt{2}-1|]$

$=\frac{1}{\sqrt{2}}\left[\log (\frac{\sqrt{2}+1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1})\right]$

$=\frac{1}{\sqrt{2}}\log \frac{(\sqrt{2}+1{)}^2}{2-1}=\frac{1}{\sqrt{2}}\log \frac{(\sqrt{2}+1{)}^2}{1}$

$=\sqrt{2}\log (\sqrt{2}+1)$

Hence, $I=\sqrt{2}\log (\sqrt{2}+1)$