How do I integrate $\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}$ How do I evaluate this indefinite integral, for $|k| < 1$:
$$
\int\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}\mathrm{d}x
$$
I tried the change of variable $t=\sin x$, and obtained two integrals, but I can't integrate either. 

Maple says:
$$
\int \!{\frac {\sqrt {1-{k}^{2}  \sin^2 \left( x \right) 
 }}{\sin \left( x \right) }}{dx}
= \frac{1}{2}\left[k\ln  \left( 2 \right) +k
\ln  \left( k \right) -k\ln  \left( -2\, \cos^2 \left( x \right) 
 {k}^{2}+2\,\cos \left( x \right)k \sqrt { \cos^2
 \left( x \right){k}^{2}-{k}^{2}+1}+{k}^{2}-1 \right) -
{\rm atanh} \left({\frac { \cos^2 \left( x \right) 
 {k}^{2}+ \cos^2 \left( x \right)-{k}^{
2}+1}{2\cos \left( x \right) \sqrt {  \cos^2 \left( x \right) 
 {k}^{2}-{k}^{2}+1}}} \right) \right]
$$
