Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 8

Evaluate $\displaystyle \int x^2 \cos mx dx$ by using Integration by parts.
If we let $u = x^2$ and $dv = \cos mx dx$, then
$du = 2x dx$ and $\displaystyle v = \int \cos mx dx = \frac{1}{m} \sin mx$

So,

$
\begin{equation}
\begin{aligned}
\int x^2 \cos mx dx = uv - \int vdu &= \frac{x^2}{m} \sin (mx) - \int \left( \frac{1}{m} \sin mx \right) (2x dx)\\
\\
&= \frac{x^2}{m} \sin (mx) - \frac{2}{m} \int x \sin (mx) dx
\end{aligned}
\end{equation}
$


To evaluate $\displaystyle \int x \sin (mx) dx$, we must use integration by parts once more, so...
If we let $u_1 = x$ and $dv_1 = \sin (mx) dx$, then
$ du_1 = dx$ and $\displaystyle v_1 = \int \sin (mx) dx = \frac{1}{m} \left(-\cos (mx) \right)$

Thus,

$
\begin{equation}
\begin{aligned}
\int x \sin (mx) dx &= u_1 v_1 - \int v_1 du_1 = \frac{-x}{m} \cos (mx) - \int \frac{-\cos (mx) dx}{m}\\
\\
&= \frac{-x \cos (mx)}{m} + \frac{\sin(mx)}{m^2} + c
\end{aligned}
\end{equation}
$


Therefore,



$
\begin{equation}
\begin{aligned}
\int x^2 \cos mx dx &= \frac{x^2}{m} \sin (mx) - \frac{2}{m} \left[ \frac{-x \cos (mx)}{m} + \frac{\sin (mx)}{m^2} + c\right]\\
\\
&= \frac{x^2 \sin (mx)}{m} + \frac{2x \cos (mx)}{m^2} - \frac{2 \sin (mx)}{m^3} + c
\end{aligned}
\end{equation}
$