Question

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

Answer 1

To prove:- ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{x}{\sin x+\cos x}dx=\frac{\pi }{2\sqrt{2}}\log (\sqrt{2}+1)$

Solution:-

$I={\int }_0^{\frac{\pi }{2}}\frac{x}{\sin x+\cos x}dx.....\left(1\right)$

$I={\int }_0^{\frac{\pi }{2}}\frac{(\frac{\pi }{2}+0)-x}{\sin (\frac{\pi }{2}+0-x)+\cos (\frac{\pi }{2}+0-x)}dx$

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

Adding equation $\left(1\right)\&\left(2\right)$, we have

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

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

$I=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}\frac{dx}{\sin x+\cos x}$

$I=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{2}(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x)}$

$=\frac{\pi }{4\sqrt{2}}{\int }_0^{\frac{\pi }{2}}\frac{dx}{\sin \frac{\pi }{4}\sin x+\cos \frac{\pi }{4}\cos x}$

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

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

$=\frac{\pi }{4\sqrt{2}}[\log (\sec (\frac{\pi }{2}-\frac{\pi }{4})+\tan (\frac{\pi }{2}-\frac{\pi }{4}))-\log (\sec (0-\frac{\pi }{4})+\tan (0-\frac{\pi }{4}))]$

$=\frac{\pi }{4\sqrt{2}}[\log (1+\sqrt{2})-\log (\sqrt{2}-1)]$

$=\frac{\pi }{4\sqrt{2}}\left(\log \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)\right)$

$=\frac{\pi }{4\sqrt{2}}\left(\log \left(\frac{{(\sqrt{2}+1)}^2}{2-1}\right)\right)$

$=\frac{\pi }{4\sqrt{2}}\left(2\log (\sqrt{2}+1)\right)$

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

Hence proved.