Prove that $\int_{-1}^{1} 4x \sqrt{1-x^{2}}\, dx = -3\pi$

Question

What is of interest is the assertion $$\int_{-1}^{1} 4x \sqrt{1-x^{2}}\,dx = -3\pi.$$

Since $$\pi = 2\int_{-1}^{1}\sqrt{1-x^{2}}\,dx,$$ i.e. the area of a unit circle, it suffices to prove that $$\int_{-1}^{1}x\sqrt{1-x^{2}}\,dx = -\frac{3}{2}\int_{-1}^{1}\sqrt{1-x^{2}}\,dx.$$ But I did not see the left-hand side can be equal to the right-hand side?

Answer 1

The integrand $4 x \sqrt{1 - x^2}$ is odd and the domain $[-1, 1]$ of integration is symmetric around the origin, so by symmetry the integral is zero: $$\int_{-1}^1 4x \sqrt{1 - x^2} \,dx = 0.$$

Answer 2

Hint: $u^2 = 1-x^2$. Use substitution.

Answer 3

Hint: You could try the following substitution : $x=cos(u)$ or $x=sin(u)$

Answer 4

$x=\sin \theta \implies dx= \cos \theta \ d\theta$

$\displaystyle\int x \sqrt{1-x^2}\ dx=\displaystyle\int \sin \theta \cos^2\theta\ d\theta$

$\cos^2 \theta=z \implies -2\sin \theta\cos \theta\ d\theta=dz$

$\therefore \displaystyle\int \sin \theta \cos^2\theta\ d\theta=-\dfrac{1}{2}\displaystyle\int \sqrt{z}\ dz$