5
$\begingroup$

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.

  • 0
    @Fabian: yeah. I did things in a hurry and I didn't wait W|A for providing with complete information.2012-09-22

2 Answers 2

4

$\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.

  • 0
    @Mercy There is always handwavingness when finding primitives, I know. But things usually work out.2012-09-22
3

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.

  • 0
    @Valentin: OK, thanks! (+1)2012-09-22