May I know how this integral was evaluated by using the theory of elliptic integrals? I can not solve the following integral using the theory of elliptic integrals:
$$\int_a^b \frac{\sin(x)}{\sqrt{c-\sin(x)}}dx$$
Where $a\geq 0, b>0, c>0$.
Wolfram$|$Alpha showed the following result:
http://www.wolframalpha.com/input/?i=integrate+%5B%2F%2Fmath:sinx%2F(a-sinx)%5E(1%2F2)%2F%2F%5D+dx
But I do not understand how Wolfra$|$Alpha came to this result.
Thanks in advance.
P.S:
How can they have also resulted in the form of hypergeometric functions ?
 A: I will show the relation to the elliptic integrals, as I said in a comment (and Albas in the next comment):
$$\sin(x)=-\sin(-x)=-\cos \left(x +\frac{\pi}{2} \right)=-1+2 \sin^2 \left( \frac{x}{2}+\frac{\pi}{4} \right)$$
$$\phi=\frac{x}{2}+\frac{\pi}{4}$$
$$x=2\phi-\frac{\pi}{2}$$
$$\alpha=\frac{a}{2}+\frac{\pi}{4}$$
$$\beta=\frac{b}{2}+\frac{\pi}{4}$$
$$\gamma=\sqrt{\frac{2}{c+1}}$$

$$\int_a^b \frac{\sin(x)}{\sqrt{c-\sin(x)}}dx=\sqrt{2} \gamma \int_\alpha^\beta \frac{-1+2 \sin^2 \phi}{\sqrt{1-\gamma^2 \sin^2 \phi}}d\phi=$$
$$=2\sqrt{2} \gamma \int_\alpha^\beta \frac{\sin^2 \phi ~d\phi}{\sqrt{1-\gamma^2 \sin^2 \phi}}-\sqrt{2} \gamma \int_\alpha^\beta \frac{d\phi}{\sqrt{1-\gamma^2 \sin^2 \phi}}$$

The second integral is incomplete elliptic integral of the first kind $F(\alpha, \gamma)-F(\beta, \gamma)$ and the second can be calculated in terms of incomplete elliptic integrals of first and second kind.

I will show the full solution for the easiest case of complete elliptic integrals.
$$\alpha=0$$
$$\beta=\frac{\pi}{2}$$
This means that:
$$a=-\frac{\pi}{2}$$
$$b=\frac{\pi}{2}$$
Now the second integral will just be:
$$ \int_0^{\frac{\pi}{2}} \frac{d\phi}{\sqrt{1-\gamma^2 \sin^2 \phi}}=K(\gamma)$$
On the other hand, the complete elliptic integral of the second kind is defined:
$$\int_0^{\frac{\pi}{2}} \sqrt{1-\gamma^2 \sin^2 \phi} ~~d\phi=E(\gamma)$$
$$\frac{d E}{d \gamma}=-\gamma \int_0^{\frac{\pi}{2}} \frac{\sin^2 \phi ~d\phi}{\sqrt{1-\gamma^2 \sin^2 \phi}}=\frac{1}{\gamma}(E(\gamma)-K(\gamma))$$
So the first integral is:
$$\int_0^{\frac{\pi}{2}} \frac{\sin^2 \phi ~d\phi}{\sqrt{1-\gamma^2 \sin^2 \phi}}=\frac{1}{\gamma^2}(K(\gamma)-E(\gamma))$$
