18
$\begingroup$

$\int \sqrt{\sin x} ~dx.$

Does there exist a simple antiderivative of $\sqrt{\sin x}$? How do I integrate it?

  • 2
    As an aside, $~\displaystyle\int_0^\tfrac\pi2\sqrt{\sin x}~dx ~=~ \int_0^\tfrac\pi2\sqrt{\cos x}~dx ~=~ 2\sqrt\pi~\frac{\Gamma\bigg(\dfrac34\bigg)}{\Gamma\bigg(\dfrac14\bigg)}~.~$ See [Wallis' integrals](http://en.wikipedia.org/wiki/Wallis'_integrals) for more details.2015-05-09

1 Answers 1

24

Since $\sqrt{\sin(x)} = \sqrt{1 - 2 \sin^2\left(\frac{\pi}{4} -\frac{x}{2}\right)}$, this matches with the elliptic integral of the second kind: $\begin{align*} \int \sqrt{\sin(x)} \, \mathrm{d} x &\stackrel{u = \frac{\pi}{4}-\frac{x}{2}}{=} -2 \int \sqrt{1-2 \sin^2(u)} \,\mathrm{d} u\\ &= -2 E\left(u\mid 2\right) + c = -2 E\left(\frac{\pi}{4}-\frac{x}{2}\middle|\, 2\right) + c \end{align*}$ where $c$ is an integration constant.

  • 0
    I fixed your delimiters. Remember: [comma for modulus, bar for parameter!](http://math.stackexchange.com/a/108659)2012-08-02