3
$\begingroup$

How do I evaluate the integral by using substitution?

$$\int (3x-2)^{20} dx $$

I'm just wanting to know the basics of how to do such problems, as this is the upcoming section for me.

What do I choose to use as a substitution? And where do I go from there?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ It helps to consult a standard textbook in calculus. :) $\endgroup$ – Amitabh Udayiman May 7 '12 at 6:18
12
$\begingroup$

Let $u=3x-2$. Then $\frac{du}{dx}=3$ and "therefore" $dx=\frac{du}{3}$. Our integral becomes $\int\frac{1}{3}u^{20}\,du$. This is a standard integral, we get $\frac{1}{63}u^{21}+C$. Finally, replace $u$ by $3x-2$.

There is an old joke that one replaces whatever is ugly by $u$, because $u$ is the first letter of the word "ugly." But in fact it is hard to give general rules for the strategy to use to attack an integral. Experience helps a lot: if you practice, after a while you recognize close relatives of problems you have already solved before.

You would use a very similar substitution if you wanted, for example, $\int\cos(17x-12)\,dx$. Just replace the ugly $17x -12$ by $u$. Note that $dx$ turns out to be $\frac{1}{17}du$.

Here is a fancier version of your problem. Find $\int(3x-6)(3x-2)^{20}\,dx$. Note that we could have solved your original problem by expanding $(3x-2)^{20}$, but that would have been a lot of work. We can also solve the new problem in the same way, but it would be. painful So let $3x-2=u$. Then $dx=\frac{du}{3}$, and $3x-6=u-4$. So our integral becomes $$\int \frac{1}{3}(u-4)u^{20}\,du.$$ Now we can multiply $(u-4)$ by $u^{20}$, getting $u^{21}-4u^{20}$. So our integral becomes $$\int \frac{1}{3}\left(u^{21}-4u^{20}\right)\,du,$$ not hard.

A last example! We want $\int x\sin(x^2)\,dx$. Substitute for the ugly inner function $x^2$, letting $u=x^2$. Then $\frac{du}{dx}=2x$, "so" $x\,dx=\frac{1}{2}du$. Substituting, we get $\int\frac{1}{2}\sin(u)\,du$, a standard integral. Remember, when we substitute, all traces of $x$ must disappear, meaning that $dx$ has to be expressed in terms of $du$. Note that in this case the $x$ in front helps, because it is almost $\frac{du}{dx}$, where $u=x^2$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Great, especially the old joke part. $\endgroup$ – Gigili May 7 '12 at 6:19
  • $\begingroup$ Might be worth mentioning explicitly, though you suggested it with your quotation marks, that when splitting the $du$ and $dx$, we've moved from mathematics to bookkeeping: it works and there's a reason why it works, but a statement like $dx=\frac{du}{3}$, read mathematically, is not strictly true. This is a confusing point for beginners that's often unsaid. $\endgroup$ – Korgan Rivera May 7 '12 at 14:55
  • $\begingroup$ @KorganRivera: Good point. Because of the nature of the question, I was concentrating on the "how to" and formal manipulations, in the style the OP might be expected to perform them in an elementary calculus course. A better but much longer approach would be to start with guessing say $(3x-2)^{21}$, differentiating, adjusting the guess, and gradually building up experience and insight about connection with Chain Rule. $\endgroup$ – André Nicolas May 7 '12 at 16:34
2
$\begingroup$

I will try to show how you can deal with the general case $$I=\int (ax+b)^n dx$$ where $n$ $a$ and $b$ are constants. Putting $u=ax+b$, we obtain $du=a dx$ i.e. $dx=\frac{1}{a} du$ . So, $$I=\int u^n \frac{1}{a}du=\frac{1}{a}\int u^n du=\frac{1}{a}\frac{u^{n+1}}{n+1}+C=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}+C$$.Hope that helps as well.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.