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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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Do you know the integral of $\arccos x$? – Nana Mar 24 '12 at 9:26
integration by parts should work see here – Raymond Manzoni Mar 24 '12 at 9:33
I can integrate that easily by parts: $\int arccosx dx = xarccosx - \int \frac{x}{\sqrt{1-x^2}} dx = xarccosx - \frac{1}{2} \int \frac {2x}{\sqrt{1-x^2}} dx = xarccosx - \frac{1}{2} \int (1-x^2)'(1-x^2)^{\frac{1}{2}} dx = xarccosx - \frac {1}{2} 2 \sqrt{1-x^2} + C = xarccosx - \sqrt{1-x^2} + C $. But I don't see how it helps me – Gabi Purcaru Mar 24 '12 at 9:34
Once again: "solve" is the wrong word here. One solves problems and one solves equations. One evaluates expressions. In particular, one evaluates integrals. – Michael Hardy Mar 24 '12 at 18:12
@MichaelHardy yes, you are right. Sorry – Gabi Purcaru Mar 24 '12 at 18:59
up vote 3 down vote accepted

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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well I did get to this point, but couldn't get further than this. The $\int \arccos x dx$ case worked because I could supply a $(1-x^2)'$ in the numerator; this one can't be tackled that way – Gabi Purcaru Mar 24 '12 at 9:55
It is worth it to notice that $$1 - \frac{1}{4} (x^3 - 3 x)^2 = \left( 1 - \frac{x^2}{4} \right) (1 - x^2)^2.$$ – Sangchul Lee Mar 24 '12 at 10:01
@sos440 this is exactly the hint I needed. Thank you! I'd accept this if you wrote it as an answer instead of comment – Gabi Purcaru Mar 24 '12 at 10:09

$x \arccos \left(\frac{x^3-3x}{2}\right)-3 \sqrt{4-x^2 }$

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Not a very enlightening answer, no? – J. M. Mar 25 '12 at 1:38

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