What I've tried for this problem is expanding it to $x^3\cos(2x) + \cos(2x)$ and then evaluating the respective functions as separate integrals. The first one uses tabular and the second one is simple u substitution. Is my procedure correct?
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3That is correct. You could also just immediately use tabular/by parts integration. – Andrey Kaipov Oct 16 '14 at 00:55
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why many likes for the problem? – Oct 16 '14 at 01:14
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what we want to do is eliminate x for the integral, thus do the integral by parts.
$$ \begin{align} \int (x^3+1)\cos (2x))\mathrm dx &=\frac{1}{2}(x^3+1)\sin2x - \int 3x^2\sin 2x\mathrm dx\\ &=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^2\cos2x-\frac{1}{2}\int 3x\cos2x\mathrm dx\\ &=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^2\cos2x-\frac{1}{4}3x\sin2x +\frac{3}{4}\int\sin 2x \mathrm dx \\ &=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^2\cos2x-\frac{1}{4}3x\sin2x -\frac{3}{8}\cos 2x +constant \end{align} $$
Tron
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The first integral must be integrated by parts 3 times to whittle down the $x^3$.
ncmathsadist
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