Substitution to evaluate $\int{\sqrt{\sin x}\cos^3x\ dx}$?

Question

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:

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

Answer 1

Hint:

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

Answer 2

The thing that usually works with trig integrals is $$u=\tan \frac{x}{2}.$$ Then you find sine and cosine and it usually just becomes a big mess of fractions..

Answer 3

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}