Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 24

Evaluate $\displaystyle \int^\pi_0 x^2 \cos x dx$ by using Integration by parts.
If we let $u = x^3$ and $dv = \cos x dx$, then
$du = 3x^2 dx$ and $\displaystyle v = \int \cos x dx = \sin x$

So,
$\displaystyle \int^\pi_0 x^3 \cos x dx = uv - \int v du = x^3 \sin x - \int 3x^2 \sin x dx$
To evaluate $\displaystyle \int 3x^2 \sin x dx$, we must also use integration by parts so


if we let $u_1 = 3x^2$ and $dv_1 = \sin x dx$, then
$du_1 = 6x dx$ and $\displaystyle v_1 = \int \sin x dx = - \cos x$

So,

$
\begin{equation}
\begin{aligned}
\int 3x^2 \sin x dx = u_1 v_1 - \int v_1 du_1 &= -3x^2 \cos x - \int - 6x \cos x dx\\
\\
&= -3x^2 \cos x + 6 \int x \cos x dx
\end{aligned}
\end{equation}
$

To evaluate $\displaystyle \int x \cos dx$, we let

$
\begin{equation}
\begin{aligned}
u_2 &= x \text{ and } dv_2 = \cos x dx, \text{ then}\\
\\
du_2 &= dx \text{ and } v_2 = \sin x
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
\text{so } \int x \cos x dx = u_2 v_2 - \int v_2 du_2 &= x \sin x - \int \sin x dx\\
\\
&= x \sin x - (-\cos x)\\
\\
&= x \sin x + \cos x
\end{aligned}
\end{equation}
$


Going back to the first equation,

$
\begin{equation}
\begin{aligned}
\int^\pi_0 x^3 \cos x dx &= x^3 \sin x - \int 3x^2 \sin x dx\\
\\
&= x^3 \sin x - \left[ -3x^2 \cos x + 6 \left( x \sin x + \cos x \right)\right]\\
\\
&= x^3 \sin x + 3x^2 \cos x - 6 x \sin x - 6 \cos x
\end{aligned}
\end{equation}
$


Evaluating from 0 to $\pi$,
$ = - 3 \pi^2 + 12$