<s>[INST] How to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$. I am trying to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$ and I have worked this problem a couple times and keep getting the same answer. So I will show my process and please point my errors out. 
$$a=\sqrt{7}\hspace{10pt} x=\sqrt{7}\sin\theta\hspace{10pt}dx=\sqrt{7}\cos\theta d\theta \hspace{7pt}7-x^2=7\cos^2\theta$$
$$\int\frac{7\sin^2\theta \sqrt{7}\cos\theta d\theta}{\sqrt{7\cos^2\theta}}$$
$$7\int\sin^2\theta d\theta\rightarrow \frac{7}{2}\int1-\cos(2\theta)d\theta$$
$$\frac{7}{2}\left(\theta-\frac{\sin(2\theta)}{2} \right) \rightarrow \frac{7}{2}\left(\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-\frac{2\sin\theta\cos\theta}{2} \right)$$
$$\frac{7}{2}\left(\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-\frac{x}{\sqrt{7-x^2}}\frac{\sqrt{7-x^2}}{\sqrt{7}} \right)$$
$$\frac{7}{2}\left(\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-\frac{x}{\sqrt{7}} \right)$$
But apparently the answer is 
$$\frac{1}{2}\left(7\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-x\sqrt{7-x^2} \right)$$
So how do I go about getting that answer? Thanks for all the help in advance.
 [/INST] You look like you are on the right path, and have the triangle drawn out correctly. I recommend using the substitution $\frac{x}{\sqrt{1-x^2}}=\tan(\theta)$. This should convert your integral into $$7\int \! \tan(\theta)\sin(\theta)\cos(\theta) \, \mathrm{d}\theta=7\int \!\sin^2(\theta) \, \mathrm{d}\theta$$ Now using the identity $sin^2(\theta)=\frac{1-\cos(2\theta)}{2}$ our integral becomes much easier. It should be obvious that $$7\int \! \frac{1-\cos(2\theta)}{2} \, \mathrm{d}\theta=\frac{7}{2}\left[\theta-\frac{1}{2}\sin(2\theta)+C \right]=\frac{7}{2}\left[\theta-\sin(\theta)\cos(\theta)+C \right]$$ Now we can go back to our substitutions from earlier and solve for $\theta$. Using the substitution $x=\sqrt{7}\sin(\theta)$ and $\sqrt{\frac{7-x^2}{7}}=\cos(\theta)$ it shouldn't be too difficult to get $\theta=\arcsin(\frac{x}{\sqrt{7}})$ and $\theta=\arccos\left(\sqrt{\frac{7-x^2}{7}}\right)$. Now we may plug the appropriate $\theta$ back into our indefinite integral to find $$\frac{7}{2}\left[\arcsin\left(\frac{x}{\sqrt{7}}\right)-\sin\left(\arcsin\left(\frac{x}{\sqrt{7}}\right)\right)\cdot\cos\left(\arccos\left(\sqrt{\frac{7-x^2}{7}}\right)\right)+C \right]$$ $$=\frac{7}{2}\left[\arcsin\left(\frac{x}{\sqrt{7}}\right)-\left(\frac{x}{\sqrt{7}}\right)\cdot\left(\sqrt{\frac{7-x^2}{7}}\right)+C \right]$$ This should be identical to your result above, once you factor in the ${7}$ out front.
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