<s>[INST] How to find the value of this integral? $\int_0^1x^2\sqrt{1-x^2}\log(1-x)dx=-\frac{G}{4}+\frac{1}{24}+\frac{\pi}{64}-\frac{\pi}{16}\ln2$ $$I=\int_0^1x^2\sqrt{1-x^2}\log(1-x)dx=-\frac{G}{4}+\frac{1}{24}+\frac{\pi}{64}-\frac{\pi}{16}\ln2$$
 Where G is the Catalan's contant
 My try:
 Integrating by parts, we have:
 $$3I=\int_0^1x^4\ln(1-x)\frac{dx}{\sqrt{1-x^2}}+\int_0^1x^3\sqrt{1-x^2}\frac{dx}{1-x}=S+T$$
 $$\frac{x^3}{1-x}=-x^2-x-1+\frac{1}{1-x}$$
 It is easy to deduce:  $$\int_0^1x^2\sqrt{1-x^2}dx=\frac{\pi}{16},\int_0^1x\sqrt{1-x^2}dx=\frac{1}{3},\int_0^1\sqrt{1-x^2}dx=\frac{\pi}{4},\int_0^1\sqrt{1-x^2}\frac{dx}{1-x}=\frac{\pi}{2}+1$$
 $$3I=S+\frac{3\pi}{16}+\frac{2}{3}$$
 But how to deduct S?
 [/INST] It is easy to check that $\int_{0}^{1}x^2\sqrt{1-x^2}\,dx=\frac{\pi}{16}$. By setting $x=\cos\theta$ the problem boils down to computing
$$ \int_{0}^{\pi/2}\cos^2(\theta)\sin^2(\theta)\log\sin^2\frac{\theta}{2}\,d\theta$$
that is simple through the Fourier series of $\log\sin$ and $\log\cos$.
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