<s>[INST] Substitution to evaluate $\int{\sqrt{\sin x}\cos^3x\ dx}$? I'm already solving more daunting exercises, but for some reason I can't tackle this one:
$$
\int{\sqrt{\sin x}\cos^3x\ dx}
$$
None of the "obvious" substitutions ($\sin x$, $\sqrt{\sin x}$, $\cos x$, $\cos^3x$) seem to make sense. I'm probably missing something obvious. Can someone give a nudge please?
Solution based on the hints below:

*

*$\int\sqrt{\sin x }\cos^3x\ dx = \int{\sqrt{\sin x}(1-\sin^2{x}})\cos x\ dx$

*Now if we substitute $$u=\sin x \\ du=\cos x\ dx$$ We'll get: $$\int{u^{1/2}(1-u^2)\ du} = \int{u^{1/2}}\ du - \int{u^{5/2}}\ du = \frac{2}{3}u^{3/2} - \frac{2}{7}u^{7/2} + c$$

*This results in $$\frac{2}{3}\sin^{3/2}x - \frac{2}{7}\sin^{7/2}x + c$$
 [/INST] Another trivial substitution is $u^2=\sin x$.
So we have
$$\cos x\,\mathrm{d}x=2u\,\mathrm{d}u$$
As a result,
\begin{align}
\int \sqrt{\sin x}\cos^3 \,\mathrm{d}x 
&= \int \sqrt{\sin x}\, \cos^2 x \,\cos x \, \mathrm{d}x \\
&= \int \sqrt{u^2}\, (1-u^4) \,2u\, \mathrm{d}u \\
&= 2\int u^2(1-u^4)\,\mathrm{d}u\\
&= \frac{2}{3}u^3 -\frac{2}{7} u^7 + c\\
&= \frac{2}{3}\sin^{3/2}x -\frac{2}{7} \sin^{7/2}x + c
\end{align}
</s>