18
$\begingroup$

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

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

$\endgroup$
  • 7
    $\begingroup$ 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. $\endgroup$ – Clive Newstead Aug 1 '12 at 18:23
  • 2
    $\begingroup$ 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. $\endgroup$ – Lucian May 9 '15 at 19:02
24
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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