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

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

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It does not have an elementary antiderivative. If calculating this integral is a step in a wider problem, it might be worth posting the wider problem, because it's likely that you've gone wrong somewhere in your working. – Clive Newstead Aug 1 '12 at 18:23
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 for more details. – Lucian May 9 at 19:02

1 Answer 1

up vote 21 down vote accepted

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.

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I fixed your delimiters. Remember: comma for modulus, bar for parameter! – J. M. Aug 2 '12 at 2:58

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