Real Methods to Evaluate $2 \int_{-1}^{1}x^2 \sqrt{1-x^2}dx$

Question

I was recently contacted by a friend to find the values of the two following integrals by any means.

$$ I=2\int_{-1}^{1}x^2 \sqrt{1-x^2}dx$$

$$ J=\int_{-1}^{1}(1-x^2) \sqrt{1-x^2}dx$$

The first thought that I had was to do trig substitution, but I felt that would be somewhat messy. So I turned to contour integration and used the dog-bone contour $C$ (where the circles of radius $\epsilon$ are centered at -1 and 1).

For $I$ this gave:

$$\begin{align*} 0=\lim_{\epsilon\rightarrow0}\lim_{R\rightarrow\infty}\oint_{C}z^2 \sqrt{1-z^2}dz&=I+2\pi i \cdot Res_{z=\infty}\left ( z^2\sqrt{1-z^2}\right ) \\ &=I+2\pi i \cdot Res_{z=0}\left ( -\frac{\sqrt{1-1/z^2}}{z^4}\right ) \\ &=I-\frac{\pi}{4} \end{align*}$$

Which implies that $I=\pi/4$.

Similarly one can find that $J=3\pi/8$.

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I was satisfied with this result but I was disappointed that I couldn't think of any nice real methods to evaluate $I$ and $J$.

So now to my question, are there any methods of evaluation using real techniques besides trigonometric substitution that can give the values of $I$ and $J$?

I'm sure there are, I just can't think of any.

Answer 1

In order to compute both integrals you may exploit: $$ \forall a,b>0,\quad \int_{0}^{1} x^{a-1}(1-x)^{b-1}\,dx = B(a,b) = \frac{\Gamma(a)\,\Gamma(b)}{\Gamma(a+b)}, $$ where: $$ \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi},\qquad \Gamma(z+1)=z\,\Gamma(z).$$

Answer 2

hint:$I = 4\displaystyle \int_{0}^1 x^2\sqrt{1-x^2}dx$. Let $x = \sin \theta, 0 \leq \theta \leq \dfrac{\pi}{2}$. Can you continue ? the same trick used for evaluating $J$.

Answer 3

Hint: Change the variables: let $\sin t = \sqrt{1-x^2}$, $\cos t =x$ (then $dx=-\sin t \, dt$)... And use the Trigonometric integration ...