I'm trying to find out the antiderivative. My approach is:
$\int \sqrt{\csc x-\sin x} dx = \int \sqrt{\frac{1}{\sin x}-\sin x} dx= \int \sqrt{\frac{1-\sin^2x}{\sin x}}dx $
Then: $\int \sqrt{\frac{\cos^2 x}{\sin x}} dx = \int \frac{\cos x}{\sqrt{\sin x}} dx $
Let $u$ be $\sin x$ so $\int \frac{du}{\sqrt{u}} = 2\sqrt{u} + k$
Finally $\int \sqrt{\csc x-\sin x} dx = 2\sqrt{\sin x} +k$
I thought I was right but apparently neither WolframAlpha nor the other antiderivative calculators agree with my result.
I don't know where I went wrong, I'd very much appreciate if someone could help me out.
Thanks BTW: I haven't mastered LaTeX yet, so forgive me if it is poorly formatted.
Your answer is correct and Wolfram Alpha does indeed agree with you. When I use Wolfram Alpha, I get this: $$ \int\sqrt{\csc x - \sin x} \, dx = 2\tan x \sqrt{\cos x \cot x} + C $$
Ignoring the $+C$ (and ignoring the $\sqrt{a^2} = |a|$ paradigm with trig functions, as we often do with indefinite integrals involving trig functions and substitutions, etc...) the RHS can be simplified as follows:
\begin{align} 2\tan x \sqrt{ \cos x \cot x} &= 2 \frac{\sin x}{\cos x} \sqrt{\cos x \frac{\cos x}{\sin x}}\\[0.3cm] &= 2\frac{\sin x}{\cos x}\sqrt{\frac{\cos^2 x}{\sin x}}\\[0.3cm] &= 2\frac{\sin x}{\cos x} \cdot \frac{\cos x}{\sqrt{\sin x}}\\[0.3cm] &= 2\sqrt{\sin x} \end{align}