Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 34

Evaluate $\displaystyle \int t^3 e^{-t^2} dt$ by making a substitution first, then by using Integration by parts.
If we use $ z = -t^2$, $t^2 = -z$, then $dz = -2t dt$

so,

$
\begin{equation}
\begin{aligned}
\int t^3 e^{-t^2} dt = \int t^2 \cdot t e^{-t^2} dt &= \int - z \cdot e^z \cdot \left( \frac{dz}{-z} \right)\\
\\
&= \int \frac{1}{2} ze^z dz\\
\\
&= \frac{1}{2} \int z e^z dz
\end{aligned}
\end{equation}
$


By using integration by parts, if we let $u =z $ and $dv = e^z dz$, then
$du = dz$ and $v = e^z$

$
\begin{equation}
\begin{aligned}
\frac{1}{2} \int ze^z dz = uv - v\int du &= \frac{1}{2} \left[ ze^z - \int e^z dz \right]\\
\\
&= \frac{1}{2} \left[ ze^z - e^z \right]\\
\\
&= \frac{e^z}{z} [z -1]
\end{aligned}
\end{equation}
$

but, $z = -t^2$
So, $\displaystyle \frac{e^z}{z} [z-1] = \frac{e^{-t^2}}{2} \left[ -t^2 -1 \right] + c$