$\int\frac{\sin x}{\sqrt{1-\sin x}}dx=?$ Calculate this integral $\displaystyle\int\dfrac{\sin x}{\sqrt{1-\sin x}}dx=?$  
Effort;
$1-\sin x=t^2\Rightarrow \sin x=1-t^2\Rightarrow \cos x=\sqrt{2t^2-t^4}$
$1-\sin x=t^2\Rightarrow-\cos x dx=2tdt\Rightarrow dx=\frac{2t}{\sqrt{t^4-2t^2}}dt$
$\displaystyle\int\frac{1-t^2}{t}\cdot\frac{2t}{\sqrt{t^4-2t^2}}dt=2\int\frac{1-t^2}{\sqrt{t^4-2t^2}}dt$
$\ = 2\displaystyle\int\frac{1}{\sqrt{t^4-2t^2}}dt-2\displaystyle\int\frac{t}{\sqrt{t^2-2t}}dt$
$\ = 2\displaystyle\int t^{-1}(t^2-2)^{-\frac{1}{2}}dt-2\displaystyle\int t(t^2-2t)^{-\frac{1}{2}}dt$
But after that I don't know how to continue.

Slightly different attempt to get rid of the trig functions (assuming a domain where no sign issues cause trouble):
$$\begin{array}{rl}
\displaystyle \frac{\sin x}{\sqrt{1-\sin x}} 
& \displaystyle  
= \frac{\sin x\sqrt{1+\sin x}}{\sqrt{1-\sin x}\sqrt{1+\sin x}}\\[7 pt]
& \displaystyle  
= \frac{\sin x\sqrt{1+\sin x}}{\sqrt{1-\sin^2 x}}\\[7 pt]
& \displaystyle  
= \frac{\cos x \sin x\sqrt{1+\sin x}}{\cos^2 x}\\[7 pt]
& \displaystyle  
= \frac{\cos x \sin x\sqrt{1+\sin x}}{1-\sin^2 x}
\end{array}$$
Now let $t = \sin x$:
$$\int \frac{\sin x}{\sqrt{1-\sin x}}  \,\mbox{d}x 
=\int \frac{\cos x \sin x\sqrt{1+\sin x}}{1-\sin^2 x}\,\mbox{d}x  \to 
\int \frac{t\sqrt{1+t}}{1-t^2} \,\mbox{d}t$$
This can be rationalized with $u^2 = 1+t$ to get (after simplifying):
$$\int \left( -2 - \frac{2}{u^2-2} \right) \, \mbox{d}u$$
This appears to be a longer route than the suggestions given in some other answers ;o).
