<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] I suppose that the error is when you worked $$\frac{2\sin\theta\cos\theta}{2}=\sin\theta\cos\theta$$ You had $\sin\theta=\frac{x}{\sqrt 7}$, so $\cos^2\theta=1-\frac{x^2}{ 7}$, $cos \theta=\sqrt{1-\frac{x^2}{7}}$ and then $$\sin\theta\cos\theta=\frac{x}{\sqrt 7}\sqrt{1-\frac{x^2}{7}}=\frac{1}{7}x\sqrt{7-{x^2}} $$ 
I am sure that you can take from here.
By the way, it could have been faster getting rid of the radical in denominator, setting $x=\sqrt{7-u^2}$, $dx=-\frac{u}{\sqrt{7-u^2}}$ so the integrand reduces to $-\sqrt{7-u^2}$ from which you start the substitution (just as you did).
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