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Consider the integrand

$\qquad \cos(x+\sin^{-1}(x)) = \sqrt{1-x^2} \cos(x) - x \sin(x)$

If I ask Mathematica to integrate $\cos(x+\sin^{-1}(x))$, Mathematica is unable to do so analytically (due to the contribution from $\sqrt{1-x^2} \cos(x)$), however, Mathematica is able to evaluate the integral of $x \sin(x)$. Is there any command that instructs Mathematica to evaluate an integral as much as it can? That is, if I ask Mathematica to compute the integral of $\cos(x+\sin^{-1}(x))$, I want an output that looks like:

$\qquad \int \cos(x+\sin^{-1}(x)) \mathrm dx = \int \sqrt{1-x^2} \cos(x) \mathrm dx -x \cos(x) + \sin(x)$

Such a command would be quite useful with more complicated integrands.

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closed as off-topic by m_goldberg, MarcoB, Edmund, user9660, Öskå Mar 3 '16 at 21:15

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  • $\begingroup$ Your question is quite vague and subjective. In what sense is the right integral "more integrated" than the left one? $\endgroup$ – Berg Mar 2 '16 at 23:43
  • $\begingroup$ @Berg, sorry for the ambiguity: this was a bit difficult to phrase. The right integral is "more integrated" only in that there is an analytical expression for part of it. Rather than analyzing the entire complicated left integral expression, I can break it into an easy to analyze part plus a complicated (but hopefully less so) part. $\endgroup$ – namu Mar 2 '16 at 23:48
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    $\begingroup$ A problem I see is that I don't think this can have a unique answer. A trivial example of what I mean would be something like adding 1 to the integrand on the right hand side and subtracting $x$ from the integrated function. There are all kinds of ways you can split the integrand up, and it will rarely be obvious what is the correct way to split it up. $\endgroup$ – march Mar 2 '16 at 23:57
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    $\begingroup$ I agree with march. Mathematica would have to be an artificial intelligence with a sense of mathematical esthetics to do what you ask for. And even then you would probably quarrel with over the choices it made. $\endgroup$ – m_goldberg Mar 3 '16 at 0:34
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    $\begingroup$ I'm voting to close this question as off-topic because there are no possible answers to this question. $\endgroup$ – m_goldberg Mar 3 '16 at 0:35
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f[x_] = Cos[x + ArcSin[x]];

Integrate[#, x] & /@ (f[x] // TrigExpand)

enter image description here

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