A definite integral: $\int_{0}^{\pi/2} \frac{\sin x~\mathrm dx}{\sin x+\cos x+ e^x}$

Question

Mathematica can do this integral, $$\int_{0}^{\pi/2} \frac{\sin x~ \mathrm dx}{\sin x+\cos x+ e^x}\,,$$ the question is: how to do it by hand?

Answer 1

$$\int_{0}^{\pi/2} \frac{\sin x~ dx}{\sin x+\cos x+ e^x}dx$$

$$=\int_{0}^{\pi/2} \frac{e^{-x}\sin x~ dx}{e^{-x}(\sin x+\cos x)+ 1}dx$$

Put $1+e^{-x}(\sin x+\cos x)=t$. Then, $-2e^{-x}\sin x dx=dt$.

The integral changes to

$$=\int_{2}^{1+e^{-\pi/2}} \frac{-1}{2t}dt$$

$$=\frac{1}{2}\ln\left(\frac{2}{1+e^{-\pi/2}}\right)$$

Answer 2

Rewrite $$ \frac{\sin(x)}{e^x+\sin (x)+\cos (x)}=\frac{1}{2}-\frac{1}{2}\frac{e^x+\cos (x)-\sin (x)}{e^x+\sin (x)+\cos (x)}$$ $$\int \frac{\sin(x)}{e^x+\sin (x)+\cos (x)}\,dx=\frac{x}{2}-\frac{1}{2} \log \left(e^x+\sin (x)+\cos (x)\right)$$