Betsy Rhone

2022-01-21

Integrating a product, one factor a derivative
$\int \frac{dx\left(t\right)}{dt}x{\left(t\right)}^{2}dt$

Elois Puryear

If integration by parts gave you 0, its

Dabanka4v

I’m going to assume that $\frac{x\left(t\right)}{dt}$ is a typo for $\frac{d\left(x\left(t\right)\right)}{dt}$.
Look at a specific example, say with $x\left(t\right)=\mathrm{sin}t$. Then
$\int \frac{d\left(x\left(t\right)\right)}{dt}x{\left(t\right)}^{2}dt=\int \mathrm{cos}t{\mathrm{sin}}^{2}tdt$,
a problem that you would most likely solve by making the substitution $u=\mathrm{sin}t,du=\mathrm{cos}tdt$, and integrating
$\int {u}^{2}du$.
But you don’t have to know what x(t) is to make this substitution. If you let $u=x\left(t\right)$, then
$du=\frac{d\left(x\left(t\right)\right)}{dt}dt$,
and $\int \frac{d\left(x\left(t\right)\right)}{dt}x{\left(t\right)}^{2}dt=\int {u}^{2}du=\frac{{u}^{3}}{3}+C=\frac{x{\left(t\right)}^{3}}{3}+C$

RizerMix

Also, since $\int \frac{dx\left(t\right)}{dt}x\left(t{\right)}^{2}dt=\int \frac{df\left(t\right)}{dt}dt$, where $f\left(t\right)=\frac{1}{3}x\left(t{\right)}^{3}$, we can use the Second Fundamental Theorem of Calculus to conclude that $\int \frac{dx\left(t\right)}{dt}x\left(t{\right)}^{2}dt=f\left(t\right)+C=\frac{1}{3}x\left(t{\right)}^{3}+C$. No substitution needed.