Question

Evaluate ${\int }_0^{\frac{\pi }{2}}\frac{\sin x}{1+\cos x+\sin x}dx$

Answer 1

Apply substitution u= $\tan \frac{x}{2}$

$=\int \frac{2u}{(u+1)(1+u^2)}du=2\cdot \int \frac{u}{(u+1)(1+u^2)}du=2\cdot \int \frac{u+1}{2(u^2+1)}-\frac{1}{2(u+1)}du=2(\int \frac{u+1}{2(u^2+1)}du-\int \frac{1}{2(u+1)}du)\therefore \frac{u+1}{u^2+1}=\frac{u}{u^2+1}+\frac{1}{u^2+1}\int \frac{u}{u^2+1}du=\frac{1}{2}\ln |u^2+1|\int \frac{1}{u^2+1}du=\arctan \left(u\right)=2(\frac{1}{2}(\frac{1}{2}\ln |u^2+1|+\arctan \left(u\right))-\frac{1}{2}\ln |u+1|)u=\tan \left(\frac{x}{2}\right)=2(\frac{1}{2}(\frac{1}{2}\ln |{\tan }^2\left(\frac{x}{2}\right)+1|+\arctan \left(\tan \left(\frac{x}{2}\right)\right))-\frac{1}{2}\ln |\tan \left(\frac{x}{2}\right)+1|)=\frac{1}{2}\ln \left({\sec }^2\left(\frac{x}{2}\right)\right)+\arctan \left(\tan \left(\frac{x}{2}\right)\right)-\ln |\tan \left(\frac{x}{2}\right)+1|$

Computing the boundaries

${\int }_0^{\frac{\pi }{2}}\frac{\sin \left(x\right)}{1+\cos \left(x\right)+\sin \left(x\right)}dx=-\frac{1}{2}\ln \left(2\right)+\frac{\pi }{4}-0=\frac{\pi }{4}-\frac{1}{2}\ln \left(2\right)$