Solving the indefinite integral of a trig function I'd like to ask for some feedback on my calculation.  Please let me know if you spot any mistakes in my technique:
$$\int{\frac{1}{\sqrt{x}}\sin^{-1}{\sqrt{x}}}\,\,dx$$
Using substitution:
$$u = \sqrt{x},\,\,\frac{du}{dx}=\frac{1}{2\sqrt{x}},\,\,du=\frac{1}{2\sqrt{x}}dx$$
So
$$2\int{\sin^{-1}{u}}\,\,du = u\sin^{-1}{u}+\sqrt{(1-u^2)}+c$$
$$=2\sqrt{x}\sin^{-1}{\sqrt{x}}+2\sqrt{1-x}+c$$
I'd greatly appreciate anyone's input.  Thank you!

Your calculation and technique are correct. Checking with Wolfram Alpha, we get the same answer. You could integrate $sin^{-1}(x)$ by parts, but if you were taught it as an identity, then your technique works.
