Question: 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!

Answer: 
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.
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.