Question

Integrate: $\sqrt{\sin 2x}\cos 2x$

Answer 1

$I=\int \sqrt{\sin 2x}\cos 2xdx$

$u=\sin 2x$

$du=2\cos 2xdx$

$\cos 2xdx=\frac{du}{2}$

$I=\int \sqrt{u}\frac{du}{2}$

$=\frac{1}{2}\int u^{1/2}du$

$=\frac{1}{2}\frac{u^{3/2}}{3/2}+c$

$=\frac{1}{3}{u}^{3/2}+c$

$=\frac{1}{3}(\sin 2x{)}^{3/2}+c$