$\int \sqrt{\sin x}\cos x dx=\frac{2}{3}{\left(\sin x\right)}^{3/2}+C$ is valid when $x$ lies in the interval

Given, $\int \sqrt{\sin x}\cos x dx=\frac{2}{3}{\left(\sin x\right)}^{\frac{3}{2}}+C$ 

Let $y=\int \sqrt{\sin x}\cos x dx$

Let, $\sin x=t$

Differentiating with respect to $x$, we get

$\cos xdx=dt$

$\therefore y=\int \sqrt{t} dt$

$\Rightarrow y=\frac{2}{3}{t}^{\frac{3}{2}}+C$

$\Rightarrow y=\frac{2}{3}{\left(\sin x\right)}^{\frac{3}{2}}+C$

But this is valid only when $\sqrt{\sin x}$ is defined.

$\therefore \sin x>0$

$\therefore x\in \left(2n\pi ,\left(2n+1\right)\pi \right), n\in \mathrm{\mathbb{N}}$