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I'm solving a differential equation and have to integrate this term:

$\int \frac{dx(t)}{dt} x(t)^2 dt$

Partial integration gave me $0$ as result, so I gave it a try on wolframalpha. This came up with a solution that is analog to results I've seen for similar equations (The derivative just seems to equal to $1$ and we use a usual integration).

Wolframalphas solution: $\int \frac{dx(t)}{dt} x(t)^2 dt = \frac{x(t)^3}{3} + c$

My problem is, I can't follow the steps wolframalpha shows.

Why can you substitute like wolframalpha does? What are the rules used to do that?

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    Oh, yes, sorry. I corrected it. x at time t is meant by x(t).2012-04-09

3 Answers 3

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If integration by parts gave you $0$, it's probably because you integrated wrong.

The simplest way to do this problem is by substitution: let $u=x(t)$. Then du = x'(t)\,dt = \frac{dx}{dt}\,dt.

So we can do a simple substitution: $\int\frac{dx(t)}{dt}(x(t))^2\,dt = \int u^2\,du = \frac{1}{3}u^3+C = \frac{1}{3}(x(t))^3 + C.$

As for integration by parts, if we set $u=(x(t))^2$, $dv = \frac{dx(t)}{dt}\,dt$, then we can take $v=x(t)$, du = 2x(t)x'(t)\,dt, so we would get $\int\frac{dx(t)}{dt}(x(t))^2\,dx = (x(t))^3 - \int 2(x(t))^2\frac{dx(t)}{dt}\,dt.$ Note that the integral on the right is the same as the integral on the left, but multiplied by $-2$; if we move it to the left hand side, we obtain $\int\frac{dx(t)}{dt} (x(t))^2\,dt + 2\int\frac{dx(t)}{dt}(x(t))^2|,dt = (x(t))^3+C.$ Now adding the two integrals and dividing by three we get $\begin{align*} 3\int\frac{dx(t)}{dt}(x(t))^2\,dt &= (x(t))^3 + C\\ \int\frac{dx(t)}{dt}(x(t))^2\,dt &= \frac{1}{3}(x(t))^3 + c. \end{align*}$

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    @MarcoSchulte: That's how substitution and differentials work. If you set $u=\sin t$, then $du = d(\sin t) = \frac{d(\sin t)}{t}dt = \cos t\,dt$. Here, $x(t)$ is the name of a function. So if $u=x$, then $du=dx$, and $dx = \frac{dx}{dt}\,dt = x'(t)\,dt$.2012-04-09
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I’m going to assume that $\dfrac{x(t)}{dt}$ is a typo for $\dfrac{d(x(t))}{dt}$.

Look at a specific example, say with $x(t)=\sin t$. Then $\int\frac{d(x(t))}{dt}x(t)^2dt=\int\cos t\,\sin^2tdt\;,$ a problem that you would most likely solve by making the substitution $u=\sin t$, $du=\cos tdt$, and integrating $\int u^2 du\;.$

But you don’t have to know what $x(t)$ is to make this substitution. If you let $u=x(t)$, then $du=\frac{d(x(t))}{dt}dt\;,$ and $\int\frac{d(x(t))}{dt}x(t)^2dt=\int u^2 du=\frac{u^3}3+C=\frac{x(t)^3}3+C\;.$

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    Oh, damn it. Yes, it is clear now. Of course $du$ IS the derivative of $u$. But thank you both for pushing my thoughts in the right direction!2012-04-09
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Also, since $\int \frac{dx(t)}{dt} x(t)^2 dt = \int \frac{d f(t)}{dt} dt$, 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.