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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?

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    $\begingroup$ It helps to consult a standard textbook in calculus. :) $\endgroup$ Commented May 7, 2012 at 6:18

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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 $u=x^2$.

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  • $\begingroup$ Great, especially the old joke part. $\endgroup$ Commented May 7, 2012 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. $\endgroup$ Commented May 7, 2012 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. $\endgroup$ Commented May 7, 2012 at 16:34
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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.

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