Question

The value of integral ${\int }_0^{\pi /2}\frac{{\sin }^{2}x}{\sin x+\cos x}dx$ is equal to

• A. $\sqrt{2}\left(\log \sqrt{2}\right)$
• B. $\sqrt{2}(\sqrt{2}+1)$
• C. $\log (\sqrt{2}+1)$
• D. None of the above

Answer 1

The correct option is D None of the above

Let $I={\int }_0^{\pi /2}\frac{{\sin }^{2}x}{\sin x+\cos x}dx$ ....(i)
Now, $I={\int }_0^{\pi /2}\frac{{\sin }^2(\frac{\pi }{2}-x)}{\sin (\frac{\pi }{2}-x)+\cos (\frac{\pi }{2}-x)}$
$\Rightarrow I={\int }_0^{\pi /2}\frac{{\cos }^{2}x}{\sin x+\cos x}dx$ ....(ii)
On adding equations (i) and (ii), we get
$2I={\int }_0^{\pi /2}\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x+\cos x}dx$
$={\int }_0^{\pi /2}\frac{1}{\sin x+\cos x}dx$
$=\frac{1}{\sqrt{2}}{\int }_0^{\pi /2}\frac{1}{\cos (x-\frac{\pi }{4})}dx$
$=\frac{1}{\sqrt{2}}{\int }_0^{\pi /2}\sec (x-\frac{\pi }{4})dx$
$=\frac{1}{\sqrt{2}}{\left[\log \{\sec (x-\frac{\pi }{4})+\tan (x-\frac{\pi }{4})\}\right]}_0^{\pi /2}$
$=\frac{1}{\sqrt{2}}[\log (\sqrt{2}+1)-\log (\sqrt{2}-1)]$
$=\frac{1}{\sqrt{2}}\log \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$
$=\frac{1}{\sqrt{2}}\log {(\sqrt{2}+1)}^2$
$=\sqrt{2}\log (\sqrt{2}+1)$