Question

Evaluate: $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{xdx}{\sin x+\cos x}$

Answer 1

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

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

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

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

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

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

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

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