Finding the integral of an inverse cosine function?

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I've just been having trouble with this question:

"Differentiate $xcos^{-1}x$ and hence find the integral of $cos^{-1}x$. Hint: Try using the substitution $u=1-x^2$."

Finding the derivative wasn't difficult, but I don't understand how this may correlate with the integral question.

Any help would be appreciated. Thanks

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$$ \frac{\mathrm{d}}{\mathrm{d}x}\left(x \cos^{-1}x\right) = \cos^{-1}x - \frac{x}{\sqrt{1-x^2}} $$ Let us now integrate both members and we get: $$ x\cos^{-1}x = \int \cos^{-1}x\ \mathrm{d}x - \int \frac{x}{\sqrt{1-x^2}} \mathrm{d}x $$

$$ \int \cos^{-1}x\ \mathrm{d}x = x\cos^{-1}x + \int \frac{x}{\sqrt{1-x^2}} \mathrm{d}x. $$ The last integral can now be solved with the substitution in the hint you mentioned!