Question

By using the properties of the definite integral, evaluate the integral:

${\displaystyle {\int }_0^{\frac{\pi }{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$ 

Answer 1

Let ${\displaystyle I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} dx}$     ...(i)

Replace x to ${\displaystyle (\frac{\pi }{2}-x)}$ in (i)

${\displaystyle [\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx]}$

${\displaystyle I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sin (\frac{\pi }{2}-x)}}{\sqrt{\sin (\frac{\pi }{2}-x)}+\sqrt{\cos (\frac{\pi }{2}-x)}} dx}$

${\displaystyle I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} dx}$       ...(ii)

Adding (i) and (ii), we get

${\displaystyle 2I={\int }_0^{\frac{\pi }{2}}[\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}] dx}$ 

${\displaystyle ={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}}$

${\displaystyle ={\int }_0^{\frac{\pi }{2}}dx={\left[x\right]}_0^{\frac{\pi }{2}}}$

${\displaystyle =\frac{\pi }{2}-0}$

${\displaystyle =\frac{\pi }{2}}$

⇒ ${\displaystyle I=\frac{\pi }{4}}$