How to do the integration: $\int x^2\sqrt{a^2-x^2}dx$

Question

More general, how to integrate:
$\int x^n\sqrt{a^2-x^2}dx$

Answer 1

Setting $x=a\sin\theta$, we get (assuming $a>0$) $$ \int x^n\sqrt{a^2-x^2}\;dx=\int (a\sin\theta)^na\cos\theta\sqrt{a^2(1-\sin^2\theta)}\;d\theta$$ $$ =a^{n+2}\int\sin^n\theta\cos^2\theta\;d\theta $$ This integral can be handled either using trig identities or integration by parts, though both methods will be tedious for large $n$. Once this is done, one must then use $x=a\sin\theta$ and some trig to switch back to $x$.

Answer 2

By changing variable. Let $x=a\cos u$ then $dx=-a\sin u du$ by replacing to $x$ you will get: $$\int a^2\cos^2 u (a\sin u)(-a\sin u du)$$ Then you can solve as a trigonometric integral.

Answer 3

Considering $$I_n=\int x^n\sqrt{a^2-x^2}\,dx$$ it seems natural to change variable $x=a \sin(y)$ which makes $$I_n=a^{n+2}\int\cos ^2(y) \sin ^n(y)\,dy$$ Now, we can simplify it to $$I_n=a^{n+2}\left(\int\sin ^n(y)\,dy-\int\sin ^{n+2}(y)\,dy\right)$$ which are quite standard and easy to compute by reduction (have a look here).

So, if you consider (as given in the link) that $$J_n=\int \sin^n(y)\,dy=-\frac{1}{n}\cos (y) \sin ^{n-1}(y)+\frac{n-1}n J_{n-2}$$ you should end with $$I_n=a^{n+2}\left(\cos (y) \sin ^{n-1}(y) \left(\frac{\sin ^2(y)}{n+2}-\frac{1}{n}\right)+\frac{n-1}n J_{n-2}-\frac{n+1}{n+2} J_{n}\right)$$

Answer 4

Using integration by parts... choose $u=x^{n-1}$ and $dv=\sqrt{a^2-x^2}xdx$ then you have $du=(n-1)x^{n-2}dx$ and $v=-\frac{(a^2-x^2)^{3/2}}{3}$ thus:

\begin{eqnarray} \int{x^n\sqrt{a^2-x^2}dx}&=&(x^{n-1})\left(-\frac{(a^2-x^2)^{3/2}}{3}\right)-\int{-\frac{(a^2-x^2)^{3/2}}{3}}(n-1)x^{n-2}dx\\ &=&-\left(\frac{(x^{n-1})(a^2-x^2)^{3/2}}{3}\right)+\frac{(n-1)}{3}\int{(a^2-x^2)^{3/2}}x^{n-2}dx \end{eqnarray}

Now, you choose $u=x^{n-3}$ and $dv=(a^2-x^2)^{3/2}xdx$ and repeat the process as much as $n$ admit.

For $n=2$ you will have:

\begin{eqnarray} \int{x^2\sqrt{a^2-x^2}dx}&=&-\left(\frac{(x)(a^2-x^2)^{3/2}}{3}\right)+\frac{1}{3}\int{(a^2-x^2)^{3/2}}dx \end{eqnarray}

The last integral it solves by trigonometric substitution, Have fun :).