Question

Evaluate : $\int \sqrt{1+\sin x}dx$

Answer 1

$\int \sqrt{1+\sin x}dx$

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

$\left[\begin{array}{c}\because {\sin }^{2}x+{\cos }^{2}x=1\\ \sin x=2\sin \frac{x}{2}\cos \frac{x}{2}\end{array}\right]$

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

$=\int \mid \sin \frac{x}{2}+\cos \frac{x}{2}\mid dx$

$=\int (\sin \frac{x}{2}+\cos \frac{x}{2})dx$

$[\because x\in (0,\pi )\Rightarrow \frac{x}{2}\in (0,\frac{\pi }{2}\left)\right]$

$=\frac{1}{\frac{1}{2}}[-\cos \frac{x}{2}+\sin \frac{x}{2}]+c$

$=-2\cos \frac{x}{2}+2\sin \frac{x}{2}+c$

Where $c$ is constant of integration
Hence, the required integration is
$-2\cos \frac{x}{2}+2\sin \frac{x}{2}+c$