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I need to solve $\int \arccos(\frac{x^3-3x}{2})dx$. I tried integration by parts by adding an x', but it didn't work. I also tried a change of variable with $\cos(t) = \frac{x^3-3x}{2}$, but that didn't get anywhere either. Could you point me in the right direction?

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    @MichaelHardy yes, you are right. Sorry2012-03-24

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integrate by parts: Let $f=\arccos\left(\frac{x^3 - 3x}{2}\right)$, g'=1. Then $g=x$ and f'= -\frac{3(x^2-1)}{2\sqrt{1-\frac{1}{4}(x^3-3x)^2}}. Thus \begin{align*} \int fg'~dx & = x\arccos\left(\frac{x^3 - 3x}{2}\right) + \int\frac{3x^3-3x}{2\sqrt{1-\frac{1}{4}(x^3-3x)^2}}~dx \\ &= x\arccos\left(\frac{x^3 - 3x}{2}\right) + \int\frac{3x(x^2-1)}{\sqrt{4-(x^3-3x)^2}}~dx \\ \end{align*}

At this point you note that $4-(x^3-3x)^2 = -(x^2-1)^2(x^2-4).$

Can you take if from here?

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    @sos440 this is __exactly__ the hint I needed. Thank you! I'd accept this if you wrote it as an answer instead of comment2012-03-24
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$x \arccos \left(\frac{x^3-3x}{2}\right)-3 \sqrt{4-x^2 }$

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    Not a very enlightening answer, no?2012-03-25