Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 42

Evaluate $\displaystyle \int x^2 \sin 2x dx$. Illustrate and check whether your answer is reasonable by gaphing both the function and its antiderivative suppose that $c = 0$.

If we use $2 = 2x$, then $\displaystyle x = \frac{z}{2}$ so $\displaystyle dx = \frac{1}{2} dz$
$\displaystyle \int x^2 \sin 2x dx = \int \left( \frac{z}{2} \right)^2 (\sin z) \left( \frac{dz}{2} \right) = \frac{1}{8} \int z^2 \sin z dz$

By using integration by parts,
If we let $u = z^2$ and $dv = \sin z dz$. Then,
$du = 2z dz$ and $v = - \cos z$

Thus,


$\begin{equation} \begin{aligned} \frac{1}{8} \int z^2 \sin z dz = uv - \int v du &= - z^2 \cos z - \int (-\cos z) ( 2z dz)\\ \\ &= -z^2 \cos z + 2 \int z \cos z dz \end{aligned} \end{equation}$

Again by using integration by parts, if we let $u_1 = z$ and $dv_1 = \cos z dz$, then
$du_1 = dz$ and $v_1 = \sin z$


$\begin{equation} \begin{aligned} \text{so, } \int z \cos z dz = u_1 v_1 - \int v_z du_1 &= z \sin z - \int \sin z dz\\ \\ &= z \sin z - (-\cos z)\\ \\ &= z \sin z + \cos z \end{aligned} \end{equation}$


Going back to the first equation,

$\begin{equation} \begin{aligned} \frac{1}{8} \int z^2 \sin z dz &= \frac{1}{8} \left[ -z^2 \cos z + 2 (z \sin z + \cos z) \right]\\ \\ &= \frac{-z^2 \cos z}{8} + \frac{z \sin z}{4} + \frac{\cos z}{4}+ c \end{aligned} \end{equation}$


but $z = 2x$, therefore,

$\begin{equation} \begin{aligned} \int x^2 \sin 2x dx &= \frac{-(2x)^2}{8} \cos (2x) + \frac{(2x) \sin (2x)}{4} + \frac{\cos 2x}{4} + c\\ \\ &= \frac{-x^2 \cos (2x)}{2} + \frac{x}{2} \sin(2x) + \frac{\cos 2x}{4} + c \end{aligned} \end{equation}$




We can see from the graph that our answer is reasonable, because the graph of the anti-derivative $f$ is increasing when $f'$ is positive. On the other hand, the graph of $f$ is decreasing when $f'$ is negative.