Computing Definite Integrals — Question 3

Question 3 - Solution

First notice a symmetry property. The function $$x^2\sqrt{1-x^2}$$ is an even function because replacing $x$ by $-x$ does not change its value. Therefore, $$\int_{-1}^{1} x^2\sqrt{1-x^2}\,dx = 2\int_{0}^{1} x^2\sqrt{1-x^2}\,dx.$$

Now focus on the integral from $0$ to $1$.

Use a trigonometric substitution. Let $$x=\sin\theta, \qquad dx=\cos\theta\,d\theta.$$

When $x=0$, $\theta=0$, and when $x=1$, $\theta=\frac{\pi}{2}$.

Substitute into the integral: $$\int_{0}^{1} x^2\sqrt{1-x^2}\,dx = \int_{0}^{\pi/2} (\sin^2\theta)(\cos\theta)(\cos\theta)\,d\theta = \int_{0}^{\pi/2} \sin^2\theta\,\cos^2\theta\,d\theta.$$

Use the identity $$\sin^2\theta\,\cos^2\theta = \frac{1}{4}\sin^2(2\theta).$$

Then $$\int_{0}^{\pi/2} \sin^2\theta\,\cos^2\theta\,d\theta = \frac14\int_{0}^{\pi/2}\sin^2(2\theta)\,d\theta.$$

Apply the identity $$\sin^2 u=\frac{1-\cos(2u)}{2}.$$

Thus, $$\frac14\int_{0}^{\pi/2}\frac{1-\cos(4\theta)}{2}\,d\theta = \frac18\left[\theta-\frac14\sin(4\theta)\right]_{0}^{\pi/2}.$$

Evaluate: $$\frac18\left(\frac{\pi}{2}-0\right) = \frac{\pi}{16}.$$

Finally, multiply by $2$ to account for the symmetry: $$\int_{-1}^{1} x^2\sqrt{1-x^2}\,dx = 2\cdot\frac{\pi}{16} = \boxed{\frac{\pi}{8}}.$$