# Integral: $\int\frac{x^2}{\sqrt{4-x^2}} dx$

So, I am really just having one issue with this integral, but I will go through the steps I have taken.

Consider $$\int\frac{x^2}{\sqrt{4-x^2}} dx$$

First, I set $x = 2\sin\theta$, found $dx = 2\cos\theta\space d\theta$ and plugged this back in, making $$4\int \frac{\sin^2\theta}{\sqrt{1-\sin^2\theta}} \cos\theta \space d\theta$$

Using $\sin^2\theta + \cos^2\theta = 1$, I canceled $\sqrt{1-\sin^2\theta}$ with $\cos\theta$, leaving $$4\int\sin^2\theta\space d\theta$$

Using $\sin^2\theta = \frac{1}{2}(1 -\cos2\theta)$, I simplified, ending up with $$4(\frac{\theta}{2} - \frac{\sin 2\theta}{4})$$

When trying to put the solution back in terms of $x$, $\sin 2\theta$ is giving me trouble. Using the relationship from before, I got that $\theta = \sin^{-1}(\frac{x}{2})$. I am able to obviously substitute this back in for $\theta$, but again I am lost what to do with $\sin 2\theta$.

Just use that $$\sin 2\theta=2\sin \theta\cos\theta=2\left(\frac x2\right)\left(\sqrt {1-\frac {x^2}{4}}\right) =\frac {x\sqrt {4-x^2}}{2}$$
Hint: $\sin(2\theta)=2\sin(\theta)\cos(\theta)$. Use this to get everything in terms of $\theta$ instead of $2\theta$.