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!

Instead of memorizing some formula, you can use IBP to conclude your answer. That comes more intuitively to someone in my opinion.
Take $\arcsin(u)$ as the first function and $1$ as the second function. Now, using IBP,
$$I=\int\arcsin(u)\,\mathrm du=\left(\arcsin(u)\int\,\mathrm du\right)-\int\left(\frac{\mathrm d}{\mathrm dx}\left(\arcsin(u)\right)\int\,\mathrm du\right)\,\mathrm du\\ = u\arcsin(u)-\int\frac{u}{\sqrt{1-u^2}}$$
Now, make the substitution $1-u^2=t$ and $(-2u)\,\mathrm du = \mathrm dt$ to get,
$$I=u\arcsin(u)+\int\frac{\,\mathrm\,dt}{\sqrt{t}}=u\arcsin(u)+2\sqrt{t}+C$$
where $C$ is the constant of integration.
Now, completely rewrite $I$ in terms of $u$ to get the "identity" you were taught and then you proceed as you did in your own solution.
