Question

Solve:$\sqrt{\sin x}dx$

Answer 1

$I=\int \sqrt{\sin x}dx=\int \sqrt{2{\cos }^2\left(\frac{2x-\pi }{4}\right)-1}dx=\int \sqrt{1-2{\sin }^2\left(\frac{2x-\pi }{4}\right)dx}Let,u=\frac{2x-\pi }{4}\Rightarrow dx=2du\therefore I=2\int \sqrt{1-2{\sin }^{2}udu}=2E\left(\frac{u}{2}\right)=2E\left(\frac{2x-\pi }{8}\right)+C$