$y=\int _{0}^{\pi /2}\frac{\sin x}{\sin x+\cos x}dx….(1)$
Use king theorem of definite integral
$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$
\[y=\int _{0}^{\pi /2}\frac{\sin \left( \frac{\pi }{2}-x \right)}{\sin \left( \frac{\pi }{2}-x \right)+\cos \left( \frac{\pi }{2}-x \right)}dx\]
\[y=\int _{0}^{\pi /2}\frac{\cos x}{\sin x+\cos x}dx…..(2)\]
Adding equation (1) and (2)
$2y=\int _{0}^{\pi /2}\frac{\sin x}{\sin x+\cos x}dx+\int _{0}^{\pi /2}\frac{\cos x}{\sin x+\cos x}dx$
$2y=\int _{0}^{\pi /2}\frac{\sin x+\cos x}{\sin x+\cos x}dx$
$2y=\int _{0}^{\pi /2}1dx$
$2y=(x)_{0}^{\pi /2}$
$y=\frac{\pi }{4}$