How do I evaluate $\int \frac{x^4}{\sqrt{4-x^2}}dx$ using trig substitution? 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!

It should be$$ 16\int sin^4\theta d\theta $$ instead of $$8\int sin^4\theta d\theta$$
And after integration :
$$\implies 6\theta - 4sin(2\theta) + \frac{1}{2}sin(4\theta) + C$$
Now, $$\theta = sin^{-1}\frac{x}{2}$$
Hence we can substitute this value of $$\theta$$ and get the final result
