Worksheet for Section 7.1

Secton 7.1 is about Substitution, the most important of the main techniques for evaluating antiderivatives. The main idea for Substitution is to construct an antiderivative rule or technique which uses the Chain Rule for derivatives backwards, the way the other rules in Section 4.1 used derivative rules backwards. The goal is to be able to find antiderivatives for functions formed as compositions of other functions. However, because of the form of the Chain Rule, the antiderivatives will require a bit of pattern-matching to make the technique work out. Recall the Chain Rule for derivatives:

$$\frac{d}{dx}F(g(x))=F'(g(x)\cdot g'(x)$$

Just writing this backwards as an antiderivative rule, you would get something like this:

$$\int F'(g(x))\cdot g'(x)\, dx=F(g(x))+C$$

So if $F(x)$ is an antiderivative for $f(x)$ (i.e. $F'(x)=f(x)$), this can be written as follows:

$$\int f(g(x))\cdot g'(x)\, dx=F(g(x))+C$$

So the key to being able to use this technique is to recognize when an integrand has the form $f(g(x))\cdot g'(x)$, where $f(x)$ is a function for which you know an antiderivative. Here are some problems where the pattern is evident:

$$\int 2x(x^2+1)^3\, dx\qquad \int 2x\sqrt{x^2+1}\, dx\qquad \int \sin^3 x\, \cos x\, dx$$

$$\int 3\cos(3x)\, dx\qquad \int 2xe^{x^2}\, dx\qquad \int e^{\tan x}\sec^2 x\, dx$$

Constant multiples can be manipulated pretty freely in antiderivatives in order to find the right match for the $f(g(x))\cdot g'(x)$ pattern. Here are a few problems that require changing constant multiples:

$$\int x(x^2+1)^3\, dx\qquad \int x\sqrt{x^2+1}\, dx\qquad \int \sqrt{3x+2}\, dx$$

$$\int \frac{x^2}{x^3+1}\, dx\qquad \int e^{3x}\, dx\qquad \int \sin(5x)\, dx\qquad$$

Even this, though, is not quite enough to fully capture how Substitution can be used in evaluating antiderivatives. The key really is the function $g(x)$, and the appearance of $g'(x)$ in the integrand. This is often the hardest part of the pattern to recognize. In the Substitution formula above, note that if $u=g(x)$, then $$\frac{du}{dx}=g'(x)$$ -- in ``differential form,'' $du=g'(x)\, dx$. Rewriting the above Substitution formula making the change of variables $u=g(x)$ and substituting $du$ for $g'(x)\, dx$ as well, you get the following formula:

$$\int f(u)\, du=F(u)+C$$

This may not seem very enlightening, but thinking of Substitution problems this way has the effect of separating the steps in finding an antiderivative by Substitution. Look back at some of the above problems, and work them out this way. More to the point, though, this ``change of variables'' way of doing Substitution allows you to handle some antiderivatives that don't quite match the $f(g(x))\cdot g'(x)$ pattern, as in the next few problems:

$$\int x(2x+1)^5\, dx\qquad \int x\sqrt{2x+3}\, dx\qquad \int x^2\sqrt{2x+3}\, dx\qquad \int \sin^3 (5x)\, \cos (5x)\, dx$$

Don't forget, after you evaluate an antiderivative using a change of variables like this, that you have to write your answer in terms of the original variable, so you have to substitute back the $u=g(x)$ into the expression.

There are two ways to handle definite integrals when you evaluate the antiderivative by Substitution. The first is to evaluate the antiderivative using the procedure above, arriving at a final answer in terms of $x$, then plugging in the endpoints as in the Fundamental Theorem. However, by thinking of Substitution as a change of variables, it is possible to modify the definite integral in a way that allows you to skip the back-substitution step, but will nonetheless guarantee the same answer:

$$\int_a^b f(g(x))\cdot g'(x)\, dx=\int_{g(a)}^{g(b)} f(u)\, du$$

Evaluating this definite integral in terms of $u$ and plugging in these new endpoints will just give a number -- the answer for the original definite integral. To show this technique, work the first couple of problems below both ways (the antiderivatives are the same as previous problems), then do the rest just using the change of variables technique:

$$\int_0^{\pi/2} \sin^3 x\, \cos x\, dx\qquad \int_0^3 x\sqrt{2x+3}... ...d \int_0^4\frac{x}{\sqrt{2x+1}}\, dx\qquad \int_0^5\frac{x+1}{\sqrt{x+4}}\, dx$$

Here are a few more problems -- these may or may not call for Substitution:

$$\int \tan x\, dx\qquad \int \tan^2 x\, dx\qquad \int 2^{-x}\, dx\qquad \int\frac1{\sqrt{x}(\sqrt{x}+1)}\, dx$$

And here are some more examples -- in each case, evaluate the integral:

$$\int \frac{x^3}{(3x^4-5)^3}dx\qquad \int\frac{x}{\sqrt{1-4x^4}}dx\qquad \int\frac{1}{1+e^x}dx$$

$$\int\frac{1}{x^2+4x+5}dx\qquad \int\frac{x^3}{1+x}dx\qquad \int\frac{1}{x^{1/3}+x^{1/2}}dx$$

$$\int\frac{x+1}{\sqrt{9-x^2}}dx\qquad \int(\tan x)(\ln(\cos x))\, dx\qquad \int\tan^2 3x\, dx$$