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I have worked through and answered correctly the following question:

$$\int x^2\left(8-x^3\right)^5dx=-\frac{1}{3}\int\left(8-x^3\right)^5\left(-3x^2\right)dx$$ $$=-\frac{1}{3}\times\frac{1}{6}\left(8-x^3\right)^5+c$$ $$=-\frac{1}{18}\left(8-x^3\right)^5+c$$

however I do not fully understand all of what I have done or why I have done it (other than I used principles I saw in a similar example question).

Specifically I picked $-\frac{1}{3}$ to multiply the whole of the integral because it is the reciprocal of $-3$ but I do not fully understand why it is necessary to perform this step.

The next part I do not understand is on the second line what causes the $\left(-3x^2\right)$ to disappear?

Here is what I think is happening:



$$\int x^2\left(8-x^3\right)^5dx=-\frac{1}{3}\int\left(8-x^3\right)^5\left(-3x^2\right)dx$$

But I picked as stated before the reciprocal of $-3$ because it was the coefficient of the derivative of the expression $8-x^3$ not because it would leave an expression equivalent to $x^2$. For example if I alter the question slightly to:

$$\int x^3\left(8-x^3\right)^5dx$$

then by picking $-\frac{1}{3}$ the following statement would be false?

$$\int x^3\left(8-x^3\right)^5dx=-\frac{1}{3}\int\left(8-x^3\right)^5\left(-3x^2\right)dx$$


$$\int-3x^2=-3\left(\frac{1}{3}x^3\right)+c$$ $$=x^3+c$$

Which is why I am confused as to why when integrating the full question $-3x^2$ seems to disappear.

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About your last question, involving $x^3(8-x^3)^5$: Ypu cannot replace this by $-\frac{1}{3}(8-x^3)^5(-3x^2)$ because you then have changed the function. When we multiplied and divided by $-3$ in the main problem, we did not change the function. – André Nicolas Feb 27 '12 at 16:58
up vote 4 down vote accepted

There are several ways to think about the process. You are using the Substitution Rule, one version of which says $$\int g'(x)h'(g(x))\,dx = h(g(x))+ C. (\ast)$$ We can easily verify that the result is correct. Just differentiate the expression on the right-hand side. By the Chain Rule, you get $g'(x)h'(g(x))$. So the Substitution Rule $(\ast)$ is just a rewritten version of the Chain Rule.

Now look at the expression you are trying to integrate. If you let $g(x)=8-x^3$, and $h'(u)=u^5$, then what you are trying to integrate looks very much like $(\ast)$. The only difference is that $g'(x)=-3x^2$, and in our integral we instead have $x^2$.

So we wish we had $-3x^2$ instead of $x^2$. Easily done, the two differ only by multiplication by a constant. So replace $x^2$ by $-3x^2$. You have multiplied your integral by $-3$. To make sure we have done nothing to the integral, we also divide by $-3$. Thus we arrive at the integral $$\frac{1}{-3}\int (-3x^2)(8-x^3)^5\,dx.$$

Now let $g(x)=8-x^3$, and $h'(u)=u^5$. Then $h(u)=\dfrac{u^6}{6}$. Apply $(\ast)$. We get $$-\frac{1}{3} h(g(x))=-\frac{1}{3}\frac{(8-x^3)^6}{6}.$$ Finally, don't forget to add the arbitrary constant of integration.

An equivalent version of $(\ast)$ which is more generally useful is $$\int g'(x)f(g(x))\,dx =\int f(u)\,du \qquad (\ast\ast)$$ where $u=g(x)$. The evaluation goes much like before. Let $u=g(x)$, and $f(u)=u^5$. Then $g'(x)=-3x^2$, so we want $$-\frac{1}{3}\int g'(x)f(g(x)\,dx.$$
By $(\ast\ast)$ this is $$-\frac{1}{3}\int f(u)du.$$ But $$\int f(u)\,du = \int u^5\,du,$$ and now things are straightforward.

Shorthand version: Here is a very commonly used shorthand for doing essentially the same calculation. Let $u=8-x^3$. We want to express everything in terms of $u$, so that $x$ entirely disappears.

We have $\dfrac{du}{dx}=-3x^2$, "and therefore" $du =(-3x^2)dx$. It follows that $x^2\,dx=-\frac{1}{3}du$. "So" $$\int x^2(8-x^3)^5\,dx=\int -\frac{1}{3}u^5 du=-\frac{1}{18}u^6+C.$$ Finally, replace $u$ by $8-x^3$.

The above manipulation will not (and should not) make full intellectual sense to you. But it really is only a cleverly disguised version of $(\ast)$. After a while, if you get used to it, you will find that the above substitution process produces the right answer almost mechanically.

Important: Please remember that after you have found an integral, you can check easily that your answer is right, by differentiating. When you integrate, it is all too easy to be off by a minus sign, or more generally by a constant factor. Differentiating will usually reveal these kinds of slips.

A mnemonic: The $8-x^3$ is kind of ugly. So replace it by $g(x)$, or better by $u$ (which, of course, stands for "ugly"). We wish our integral had $g'(x)$, that is, $\dfrac{du}{dx}$ sitting inside it.

Well, the derivative of $u$ is sitting in our integral. OK, not quite, the derivative of $u$ is $-3x^2$ and what we have in our integral is $x^2$. Easy fix, we multiply and divide by $-3$.

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Thanks for taking the time to do this. It has helped make what I am doing clearer to me. – Aesir Feb 27 '12 at 15:18

What you are doing here is undoing the chain rule. Have you learned about the $u$-substitution formalism yet?

When you differentiate $f\circ g$, you get $f\circ g\cdot g'$. This dumps an extra factor onto the end of the expression. So, you need to soak up that extra factor to find an antiderivative. For your example, proceed as follows

You are integrating $$\int x^2(8 - x^3)^5\, dx. $$ You can see that there is a function being worked on, i.e., $x\mapsto 8 - x^3$ and another factor that is a constant multiple of its derivative, $x^2$. So we will put $u = 8 - x^3$ and $du = -3x^2\, dx$.

Now let us manipulate your integral

$$\int x^2(8 - x^3)\, dx = -{1\over 3} \int (8 - x^3)(-3x^2)\, dx = -{1\over 3} \int u^5\, du = -{1\over 18} u^6 + C = -{1\over 18}(8-u^3) + C.$$

The purpose of the $du$ formalism is to undo the multiplication of the $g'$ in the chain rule.

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Thanks ncmathsadist, your answer was very helpful as well. The soaking analogy resonated particularly well with me. – Aesir Feb 27 '12 at 15:19

You correctly recognised x^2 as "almost" thw derivative of

So put u = (8 - x^3), and find du/dx = -3x^2.

The your integral becomes (-1/3)∫(-3x2)(8−x3)^5dx

= (-1/3) ∫ u^5 (du/dx) dx = (-1/3) ∫ u^5 du -- which is rather easier to follow. It is the change of variable procedure, which is the reverse of the chain rule for derivatives.

(To verify this procedure, put I = the integral, and compare dI/dx and dI/du, using the chain rule.)

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