<s>[INST] Prove that $\int_{-1}^{1} 4x \sqrt{1-x^{2}}\, dx = -3\pi$ 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?
 [/INST] $x=\sin \theta \implies dx= \cos \theta \ d\theta$
$\displaystyle\int x \sqrt{1-x^2}\ dx=\displaystyle\int \sin \theta \cos^2\theta\ d\theta$
$\cos^2 \theta=z \implies -2\sin \theta\cos \theta\ d\theta=dz$
$\therefore \displaystyle\int \sin \theta \cos^2\theta\ d\theta=-\dfrac{1}{2}\displaystyle\int \sqrt{z}\ dz$
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