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

Beginner2022-01-21Added 30 answers

If integration by parts gave you 0, its

Dabanka4v

Beginner2022-01-22Added 36 answers

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$\int \frac{d\left(x\left(t\right)\right)}{dt}x{\left(t\right)}^{2}dt=\int {u}^{2}du=\frac{{u}^{3}}{3}+C=\frac{x{\left(t\right)}^{3}}{3}+C$

Look at a specific example, say with

a problem that you would most likely solve by making the substitution

But you don’t have to know what x(t) is to make this substitution. If you let

and

RizerMix

Skilled2022-01-27Added 437 answers

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