Question

Prove that $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^{2}x}{\sin x+\cos x}dx=\frac{1}{\sqrt{2}}{\log }_e(\sqrt{2}+1)$

Answer 1

$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^{2}x}{\sin x+\cos x}dx\cdots \left(1\right)$

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$, we have -
$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^2(\frac{\pi }{2}-x)}{\sin (\frac{\pi }{2}-x)+\cos (\frac{\pi }{2}-x)}dx$
$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\cos }^{2}x}{\sin x+\cos x}dx\cdots \left(2\right)$
On adding $\left(1\right)$ and $\left(2\right)$, we have
$2I=\frac{1}{\sqrt{2}}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{\cos (x-\frac{\pi }{4})}$
$=\frac{1}{\sqrt{2}}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sec (x-\frac{\pi }{4})dx$

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

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

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

$=\frac{1}{\sqrt{2}}{\log }_e\left|\frac{\sqrt{2}+1}{\sqrt{2}-1}\right|$

On rationalizing the denominator, we have
$2I=\frac{1}{\sqrt{2}}{\log }_e\left(\frac{(\sqrt{2}+1{)}^2}{1}\right)$
$\Rightarrow 2I=\frac{2}{\sqrt{2}}{\log }_e(\sqrt{2}+1)$
$\Rightarrow I=\frac{1}{\sqrt{2}}{\log }_e(\sqrt{2}+1)$ (proved)