Question

Evaluate:${\int }_0^{\frac{\pi }{2}}\sqrt{1+\sin x}dx$

Answer 1

We have,

$I={\int }_0^{\frac{\pi }{2}}\sqrt{1+\sin x}dx$

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

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

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

$I={[\frac{\sin \frac{x}{2}}{\frac{1}{2}}-\frac{\cos \frac{x}{2}}{\frac{1}{2}}]}_0^{\frac{\pi }{2}}$

$I=2{[\sin \frac{x}{2}-\cos \frac{x}{2}]}_0^{\frac{\pi }{2}}$

$I=2[\sin \frac{\pi }{4}-\cos \frac{\pi }{4}-(\sin 0-\cos 0\left)\right]$

$I=2[\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-0+1]$

$I=2\times 1$

$I=2$

Hence, this is the answer.