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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 ?

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    Change of variables? – 2017-01-08
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    What's the changed variable here ? – 2017-01-08
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    $3x^2\,dx = dx^3$ since $\frac{d}{dx}x^3 = \frac{dx^3}{dx}=3x^2$. – 2017-01-08
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    Thank you ! The notation he used is more clear now – 2017-01-08

2 Answers 2

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$$\int x^2 \log\sqrt{1-x^2} dx = \int x^2 \log\sqrt{1-(x^3)^{2/3}} dx$$ 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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$ \int x^2 \log \sqrt{1 - x^2} \ dx \\ = \int x^2 \log \sqrt{1 - x^2} \ \frac{dx}{dx^3} \ dx^3 \\ = \int x^2 \log \sqrt{1 - x^2} \ \frac{1}{3x^2} \ dx^3 \\ = \frac13 \int \log \sqrt{1 - x^2} \ dx^3 $