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.

• I know another substitution. Would u like to know it? :) – Mikasa Sep 7 '14 at 6:50
• Of course haha! – Kenshin Sep 7 '14 at 6:57

$$\frac{x^2}{\sqrt{7-x^2}}=\frac{7-(7-x^2)}{\sqrt{7-x^2}}=7\cdot\frac1{\sqrt{(\sqrt7)^2-x^2}}-\sqrt{(\sqrt7)^2-x^2}$$

Now, $$\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac xa$$

and $$\int\sqrt{a^2-x^2}dx=x\frac{\sqrt{a^2-x^2}}2+\frac{a^2}2\arcsin\frac xa$$

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 $$\sin\theta\cos\theta=\frac{x}{\sqrt 7}\sqrt{1-\frac{x^2}{7}}=\frac{1}{7}x\sqrt{7-{x^2}}$$
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).
• so I can't just assume my $\cos$ is $\frac{\sqrt{7-x^2}}{\sqrt{7}}$ since my triangle I am using for this has a hypotenuse of $\sqrt{7}$ adjacent side of $\sqrt{7-x^2}$ and opposite side of $x$. Sorry if I am asking a dumb question I thought that the $2$'s would cancel out from $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ and leave me with what I thought. Thanks for the help! – Kenshin Sep 7 '14 at 6:56
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.