If $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin x}{1+\sin x+\cos x}dx=k,$ then $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{1+\sin x+\cos x}$ is

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin x}{1+\sin x+\cos x}dx=k,\Rightarrow \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\cos x}{1+\sin x+\cos x}dx$=k

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{1+\sin x+\cos x}+\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin x}{1+\sin x+\cos x}dx+\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\cos x}{1+\sin x+\cos x}dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}dx$

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{1+\sin x+\cos x}=\frac{\mathrm{\pi }}{2}-2k$