Question

Evaluate : $\underset{0}{\overset{\pi /2}{\int }}\frac{xdx}{\left(\cos x+\sin x\right)}$

Answer 1

$I={\int }_0^{\pi /2}\frac{x}{cosx+sinx}dx$

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

$I={\int }_0^{\pi /2}\frac{\frac{\pi }{2}-x}{cosx+sinx}dx$

$2I={\int }_0^{\pi /2}\frac{\frac{\pi }{2}-x+x}{sinx+cosx}dx$

$I=\frac{\pi }{4}{\int }_0^{\pi /2}\frac{1}{sinx+cosx}dx$

$I=\frac{\pi }{4}\times \frac{1}{\sqrt{2}}{\int }_0^{\pi /2}\frac{1}{sinxcos\frac{\pi }{4}+sin\frac{\pi }{4}cosx}$

$I=\frac{\pi }{4\sqrt{2}}{\int }_0^{\pi /2}cosec(x+\frac{\pi }{4})dx$

$I=\frac{\pi }{4\sqrt{2}}\left[ln\right[cosec(x+\frac{\pi }{4})-cot(x+\frac{\pi }{4})]{]}_0^{\pi /2}$

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

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