Here is what I have so far:

Let $x = 2\sin\theta$, and $dx=2\cos\theta \mathrm d\theta$

then the new integral is $$\int \frac{2^4\sin^4\theta}{\sqrt{4-4sin^2\theta}}2\cos\theta \mathrm d\theta$$

Then using the identity $1-\sin^2x = \cos^2x$ we can substitute $\cos^2\theta$ and simplify,

$$8\int \sin^4\theta d\theta$$

I then use the formula of $\int \sin^nudu= -\frac{1}{n}\sin^{n-1}u\cos + \frac{n-1}{n}\int \sin^{n-2}udu$, and I end up getting $$4(\theta-\sin^3\theta \cos\theta) +C$$

Is this correct?

How do I go about re-substituting $x$ into the equation?

Thanks in advance!