Find $\int (x^3+1)(\cos(2x)) dx$ 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?
what we want to do is eliminate x for the integral, thus do the integral by parts.

$$
\begin{align}
\int (x^3+1)\cos (2x))\mathrm dx
&amp;=\frac{1}{2}(x^3+1)\sin2x - \int 3x^2\sin 2x\mathrm dx\\
&amp;=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^2\cos2x-\frac{1}{2}\int 3x\cos2x\mathrm dx\\
&amp;=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^2\cos2x-\frac{1}{4}3x\sin2x +\frac{3}{4}\int\sin 2x \mathrm dx \\
&amp;=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^2\cos2x-\frac{1}{4}3x\sin2x -\frac{3}{8}\cos 2x +constant  
\end{align}
$$
The first integral must be integrated by parts 3 times to whittle down the $x^3$.