# Integrating a product, one factor a derivative

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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Is $\frac{x(t)}{dt}$ supposed to be $\frac{dx(y)}{dt}$? Otherwise, it looks like incorrect notation. – Arturo Magidin Apr 9 '12 at 6:54
Oh, yes, sorry. I corrected it. x at time t is meant by x(t). – Marco Schulte Apr 9 '12 at 6:56

## 3 Answers

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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Ah, indeed, there was a mistake in my integration by parts. That cleared a lot up. But it looks pretty clumsy compared to the substitution. But my problem still remains: I don't see why you can substitute like you, the other answerer or wolframalpha does. I have especially a problem accepting that you can set $du = x'(t)\,dt$. Is this some fundamental thing about integrals I haven't understood? Do I just have to accept, that you can do a "string-wise" substitution like that? – Marco Schulte Apr 9 '12 at 7:25
@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 $dx = \frac{dx}{dt}\,dt = x'(t)\,dt$. – Arturo Magidin Apr 9 '12 at 15:53

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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Thanks a lot for the answer! But as I said in the comment to the other answer, I just don't understand why you are allowed to substitute like this. – Marco Schulte Apr 9 '12 at 7:32
@Marco: Do you mean that you don’t know, for instance, why you can substitute $u=\sin t$ in $\int\cos t\,\sin t dt$? – Brian M. Scott Apr 9 '12 at 7:36
Yes, my problem is especially the $du=\cos t dt$. I don't see why you can do that. A good example of my problem would probably be: Given this ridiculous stupid integral $\int sin(x) \frac{1}{x^2} ln(x)\,dx$. Lets substitute $u = sin(x)$ and $du = \frac{1}{x^2} ln(x)\,dx$. Sweet, now we have $\int u\,du = u^2 + C = sin(x)^2 + C$. Obviously this is wrong, but I don't see the difference to the equation from my question. – Marco Schulte Apr 9 '12 at 7:51
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! – Marco Schulte Apr 9 '12 at 7:55

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.

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