<s>[INST] About an integration technique My professor wrote this : 
$\displaystyle \int  x^2\log(\sqrt{1-x^2}) \,\mathrm{d}x = \frac{1}{3}\int \log(\sqrt{1-x^2})   \, \mathrm dx^3$
Can someone explain me what is going on with the integration variable ? 
 [/INST] $$\int x^2 \log\sqrt{1-x^2} dx = \int x^2 \log\sqrt{1-(x^3)^{2/3}} dx$$
Let $u = x^3$, so $du = 3x^2 dx$ or 
$$dx = \frac{du}{3x^2}$$
so the integral becomes
$$\int \frac{x^2}{3x^2} \log\sqrt{1-u^{2/3}} du = \frac{1}{3} \int \log\sqrt{1-u^{2/3}} du \ .$$
Resubstituting $u = x^3$ yields the rather confusing expression of your professor.
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