# Evaluating $\int \cos(x) \sqrt{\sin(2 x)} dx$

Evaluate the following indefinite integral:

$$\int \cos(x) \sqrt{\sin(2 x)} dx$$

Only hint I have is from W|A that expresses the integral in terms of a hypergeometric function and it looks rather ugly. Can we solve it in a simpler way and get a nicer form? Thanks.

-
I get a WA expression that only involves elementary functions... –  Fabian Sep 22 '12 at 18:59
@Fabian: yeah. I did things in a hurry and I didn't wait W|A for providing with complete information. –  Chris's sis the artist Sep 22 '12 at 19:04

$$\begin{eqnarray*} \int {\cos x\sqrt {\sin 2x} dx} &=& \int {\cos x\sqrt {2\sin x\cos x} dx} \\ &=& \sqrt 2 \int {\sqrt {\cos x\sin x} \cos xdx} \\ \begin{cases}\sin x = u\\ \cos xdx = du\end{cases} \\ &=& \sqrt 2 \int {{u^{1/2}}{{\left( {1 - {u^2}} \right)}^{1/4}}} du \end{eqnarray*}$$

Do you know how to integrate differential binomials?

See this answer of mine. Since

$$\frac{{m + 1}}{n} + p = \frac{3}{4} + \frac{1}{4} = 1$$ is an integer, you should be able to integrate this in terms of elementary functions with the instructions provided in the answer I linked to. Letting $u^2=z$ gives

$$= \frac{{\sqrt 2 }}{2}\int {{{\left( {\frac{{1 - z}}{z}} \right)}^{1/4}}} dz$$

Now let $$\frac{{1 - z}}{z} = {m^4}$$ whence $$dz = \frac{{4{m^3}dm}}{{{{\left( {{m^4} + 1} \right)}^2}}}$$

and get

$$= 2\sqrt 2 \int {\frac{{{m^4}}}{{{{\left( {{m^4} + 1} \right)}^2}}}dm}$$ which is a treatable rational function.

-
Thanks for details (+1) –  Chris's sis the artist Sep 22 '12 at 19:01
For $\pi \le x \le 3\pi/2$ we have $\sin(2x) \ge 0$, but $\sin(x),\cos(x) \le 0$, and therefore $\cos(x)=-\sqrt{1-\sin^2x}$! –  Mercy Sep 22 '12 at 19:16
@Mercy There is always handwavingness when finding primitives, I know. But things usually work out. –  Pedro Tamaroff Sep 22 '12 at 19:19

Alternatively, rewrite $$I=\sqrt{2}\int(1-(\sin x)^2)^\frac{1}{4}(\sin{x})^\frac{1}{2}d(\sin{x})\\=\sqrt{2}\int(1-t^2)^{\frac{1}{4}}t^\frac{1}{2}dt\\=\sqrt{2}\int\left(\frac{1}{t^2}-1\right)^\frac{1}{4}tdt \\=\frac{1}{\sqrt{2}}\int\left(\frac{1}{z}-1\right)^\frac{1}{4}dz$$ Now let $$\frac{1}{z}-1=u^2$$ $$-\frac{dz}{z^2}=2udu$$ $$dz=-\frac{2udu}{(1+u^2)^2}=d\left(\frac{1}{1+u^2}\right)$$ $$\sqrt{2}I=\int u^\frac{1}{2}d\left(\frac{1}{1+u^2}\right)=\frac{u^\frac{1}{2}}{1+u^2}-\int\frac{d\left(u^\frac{1}{2}\right)}{1+u^2}$$ Where the last integral is equivalently $\int\frac{dv}{1+v^4}$ to which there exist various approaches.

-
Linebreaks \\ are useful for long chains of equalities. –  Pedro Tamaroff Sep 22 '12 at 19:05
thanks, will note for future –  Valentin Sep 22 '12 at 19:06
@Valentin: OK, thanks! (+1) –  Chris's sis the artist Sep 22 '12 at 19:11