Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
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. – OFFSHARING Sep 22 '12 at 19:04

2 Answers 2

up vote 4 down vote accepted

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

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.

share|cite|improve this answer
Thanks for details (+1) – OFFSHARING 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 King 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.

share|cite|improve this answer
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) – OFFSHARING Sep 22 '12 at 19:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.