Evaluating $\int x^2 \sqrt{1-x^2}\ dx$

Question

I hope I can find a way to integrate this formula without resorting to numerical techniques.

\begin{equation} \int x^2 \sqrt{1-x^2}\ dx \end{equation}

I am not sure if there's actually a closed form for this or not? I tried integration by parts, but it seems not working! Here's my last resort before numerical solutions.

Answer 1

Hint:

Try substitution $x=\sin\theta\longrightarrow dx=\cos\theta\ d\theta$, then you will get

\begin{align} \int\sin^2\theta\cos^2\theta\ d\theta&=\int\left(\frac{\sin2\theta}{2}\right)^2\ d\theta\\[10pt] &=\frac{1}{4}\int\sin^22\theta\ d\theta\\[10pt] &=\frac{1}{4}\int\left(\frac{1-\cos4\theta}{2}\right)\ d\theta\\[10pt] \end{align}

Answer 2

Here is a solution with integration by parts and rearranging terms:

First, integrating by parts and adding and subtracting a $1$ in the numerator, $$ \begin{aligned} I=\int x^2\sqrt{1-x^2}\,dx & = \frac{x^3}{3}\sqrt{1-x^2}+\int\frac{x^3}{3}\frac{x}{\sqrt{1-x^2}}\,dx\\ &=\frac{x^3}{3}\sqrt{1-x^2}-\int\frac{x^3}{3}\sqrt{1-x^2}\,dx+\int\frac{x^3}{3}\frac{1}{\sqrt{1-x^2}}\,dx. \end{aligned} $$ We got $I$ back in the right-hand side, and moving it to the left-hand side, we find $$ I=\frac{x^3}{4}\sqrt{1-x^2}+\int\frac{1}{4}\frac{x^2}{\sqrt{1-x^2}}\,dx.\tag{1} $$ We can continue from $(1)$ in two ways. Adding and subtracting a $1$ in the numerator, we get $$ I=\frac{x^3}{4}\sqrt{1-x^2}-\frac{1}{4}\int\sqrt{1-x^2}\,dx+\frac{1}{4}\int\frac{1}{\sqrt{1-x^2}}\,dx\tag{2} $$ If we instead integrate by parts, we get $$ I=\frac{x^3}{4}\sqrt{1-x^2}-\frac{1}{4}x\sqrt{1-x^2}+\frac{1}{4}\int\sqrt{1-x^2}\,dx.\tag{3} $$ We add $(2)$ and $(3)$, and find that $$ 2I=\frac{x^3}{2}\sqrt{1-x^2}-\frac{1}{4}x\sqrt{1-x^2}+\frac{1}{4}\int\frac{1}{\sqrt{1-x^2}}\,dx, $$ and hence, finally, $$ \begin{aligned} I&=\frac{x^3}{4}\sqrt{1-x^2}-\frac{1}{8}x\sqrt{1-x^2}+\frac{1}{8}\int\frac{1}{\sqrt{1-x^2}}\,dx\\ &=\frac{x^3}{4}\sqrt{1-x^2}-\frac{1}{8}x\sqrt{1-x^2}+\frac{1}{8}\arcsin x+C. \end{aligned} $$