$I=\int\limits_{0}^{π/2}{√{\sin x}}/{√{\sin x}+√{\cos x}}dx$ ...(i) Replace $x$ to $(π/2-x)$ in (i) $\left[∵\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx\right]$$I=\int\limits_{0}^{π/2}{√{\cos x}}/{√{\cos x}+√{\sin x}}dx$ ...(ii) Adding (i) and (ii), we get$⇒2I= \int\limits_{0}^{π/2}[{√{\sin x}+√{\cos x}}/{√{\sin x}+√{\cos x}} ]dx$ $⇒2I= \int\limits_{0}^{π/2}dx=[x]^{π/2}_0$ $=π/2-0=π/2$ $⇒I=π/4$