The Substitution Rule

Section 3.2 The Substitution Rule

Note: this section is a copy of the same section from Calculus I. It’s repeated here because it is the most important of the integration rules and will be used frequently in the discussion of integration in this course. It’s worth repeating to make sure you understand it well.

Subsection 3.2.1 Differentiating Composition

Since doing integrals is doing derivatives backwards, it might be reasonable to try to reversing all the differentiation rules. Linearity works exactly the same in reverse. The inverted power rule works very similarly to the original power rule. After the power rule, thouhg, the remaining rules don’t easily reverse. For most of these rules, I’ll leave the strategies for Calculus II. But I do want to cover how to do the chain rule backwards, which is the subject of this entire week.

For a composition of functions \(f(g(x))\text{,}\) the derivative is calculated by the chain rule.

\begin{equation*} \frac{d}{dx} f(g(x)) = f^\prime(g(x)) g^\prime(x) \end{equation*}

The simplest reversal of the chain rule is to integrate the right side to produce the left.

\begin{equation*} \int f^\prime(g(x)) g^\prime(x) dx = f(g(x)) + c \end{equation*}

When I covered the chain rule, I recommended labelling the inside function with a new variable \(u = g(x)\text{.}\) This labelling becomes essential now. To implement the inverse of the chain rule, I’m essentially going to change the integral from the old variable \(x\) to a new variable \(u\) using this labelling \(u = g(x)\text{.}\)

This gives me a process, which is called the substitution rule. For a integral that looks something like the output of a chain rule, I’ll try to choose an inside function and label it \(u = g(x)\text{.}\) Then I’ll differentiate and write the following equation.

\begin{equation*} du = g^\prime(x) dx \end{equation*}

The \(dx\) in the integral is not just a piece of notation that indicates the variable. It’s actually part of the integral itself, even though it disappears in the anti-derivative process. (This \(dx\) term is, quite confusingly, called the differential.) When I do the substitution rule, I need to replace the old differential \(dx\) with the new differential\(du\text{.}\) However, I can’t just replace it directy. The previous equation gives me the appropriate replacement: I need to replace \(g^\prime(x) dx\) with \(du\text{.}\) Lastly, if there are any other piece of the original integral which aren’t accounted for in the replacements I’ve done so far, I need to figure a way to replace those pieces as well, usually going some algebra with the new variable \(u\text{.}\) The key idea is that all of the original pieces of the original variable — all of the \(x\)’s, must be replace with \(u\text{.}\) The old integral is entirely in \(x\text{;}\) the new integral is entirely in \(u\text{.}\) Once I’ve done the substitution, I get a new integral in \(u\) which is hopefuly easier to calculate than the original.

This is difficult to understand from the general description, so I’ll go direclty into a bunch of examples which show the technique.

Subsection 3.2.2 Substitution Examples - Indefinite Integrals

Example 3.2.1.

\begin{equation*} \int 2x (x^2+1)^4 dx \end{equation*}

This looks like a composition, with an inside function \(x^2 + 1\text{.}\) I’ll try a substitution \(u = x^2 +1\) to see if it makes the integral more reasonable. I need to change the differential term; here, \(u = x^2 +1\text{,}\) so \(du = (2x) dx\text{.}\) If I regroup the terms in the original integral, I can replace \(2x dx\) with \(du\text{.}\) This lets me replace all of the \(x\) variables.

\begin{equation*} \int (x^2+1)^4 (2x)dx \mapsto \int u^4 du \end{equation*}

Then can find the anti-derivative by reversing the power rule.

\begin{equation*} \int u^4 du = \frac{u^5}{5} + c \end{equation*}

After I find an antiderivative in the new variable \(u\text{,}\) I need to return to the original variable by undoing the substitution. I simply replace \(u\) with \(x^2+1\text{.}\) This gives me an antiderivative for the original function in the original variable.

\begin{equation*} \int 2x (x^2+1)^4 dx = \int u^4 du = \frac{u^5}{5} + c = \frac{(x^2+1)^5}{5} + c \end{equation*}

Example 3.2.2.

\begin{equation*} \int xe^{x^2} dx = \int (e^{x^2}) (xdx) \end{equation*}

I use the substitution \(u = x^2\text{.}\) The matching change in the differential is \(du = 2xdx\text{.}\) I’ve grouped \(xdx\) in the integral; I can replace this with \(\frac{1}{2} du\text{.}\)

\begin{equation*} \int e^{x^2} xdx = \int e^u \frac{1}{2} du = \frac{1}{2} \int e^u du \end{equation*}

After taking out the constant, I get the integral of \(e^u\text{,}\) which is an integral I know. I can finish the integral and then replace the substitution with the original function.

\begin{equation*} = \frac{e^u}{2} + c = \frac{e^{x^2}}{2} + c \end{equation*}

Example 3.2.3.

\begin{equation*} \int \frac{x}{x-2} dx \end{equation*}

Here there isn’t as obvious a substitution. However, I might be able to simplify the denominator by using the substitution \(u = x - 2\text{.}\) The matching differential substitution is simply \(du = dx\text{.}\) I also need to substitute the numerator. I can do this by adding \(2\) to each side of the substitution, giving \(u + 2 = x\text{.}\)

\begin{equation*} \int \frac{x}{x-2} dx = \int \frac{u+2}{u} du \end{equation*}

Then I can split this into two pieces, since \(\frac{u+2}{u} = \frac{u}{u} + \frac{2}{u} = 1 + \frac{2}{u}\text{.}\) After that, I can use linear to split this into two integrals. The integral of a constant is \(u\) and the integral of \(\frac{1}{u}\text{,}\) from the tables, is \(\ln |u|\text{.}\) Finally, at the end I reverse the substitution.

\begin{equation*} = \int 1 + \frac{2}{u} du = \int du + 2 \int \frac{1}{u} du = u + 2 \ln |u| + c = x - 2 + 2\ln|x-2| + c \end{equation*}

The \(-2\) could be combined with the constant \(c\text{,}\) since the constant is unknown. This isn’t necessary, but you could write the solution with slightly fewer characters with this change.

\begin{equation*} = x + 2 \ln|x-2| + c \end{equation*}

Example 3.2.4.

\begin{equation*} \int \frac{1}{10x-3} dx \end{equation*}

I use the substitution \(u = 10x - 3\text{,}\) again to simplify the denominator. The associated change to the differential is \(du = 10dx\text{,}\) which I can rearrange to \(\frac{1}{10} du = dx\text{.}\)

\begin{equation*} \int \frac{1}{10x - 3} = \int \frac{1}{u} \frac{1}{10} du \end{equation*}

I can pull out the constant. Then the antiderivative of \(\frac{1}{u}\) is known from the tables. After the antiderivative, I can reverse the substitution to return to the original variable.

\begin{equation*} \frac{1}{10} \int \frac{1}{u} du = \frac{\ln |u|}{10} + c = \frac{ \ln| 10x -3 |}{10} + c \end{equation*}

Subsection 3.2.3 Substitution Examples - Definite Integrals

When I do substitution with definite integrals, I also need to change the bounds. If \(a\) and \(b\) are the bounds in \(x\) and \(u = g(x)\) is the relationship between \(u\) and \(x\text{,}\) then \(g(a)\) and \(g(b)\) will be the bounds in \(u\text{.}\) One nice thing about definite integrals is that I can use these new bounds to evaluate the integral. I don’t have to substitute back after I finish.

Example 3.2.5.

\begin{equation*} \int_0^2 \frac{2x}{(x^2+1)^2} dx \end{equation*}

I use the substitution \(u = x^2 + 1\text{,}\) with the matching differential change \(du = 2x dx\text{.}\) Since there is a \(2x dx\) in the integral, no more algebra is necessary. I also need to change the bounds. When \(x=0\text{,}\) \(u = x^2 + 1 = 1\text{;}\) when \(x = 2\text{.}\) \(u = 2^2 + 1 = 5\text{.}\)

\begin{equation*} \int_0^2 \frac{2x}{(x^2+1)^2} = \int_1^5 \frac{1}{u^2} du \end{equation*}

Then I can write \(\frac{1}{u^2}\) as \(u^{-2}\) and use the reverse power rule to find the antiderivative. After finding the antiderivative, I evalute on the bounds (with the change in the bounds, there is no need to reverse the substitution).

\begin{equation*} = \left. -\frac{1}{u} \right|_1^5 = - \frac{1}{5} + 1 = \frac{4}{5} \end{equation*}

Example 3.2.6.

\begin{equation*} \int_{-1}^2 x^2 e^{x^3+1} dx = \int_{-1}^2 e^{x^2 + 1} (x^2 dx) \end{equation*}

I use the substitution \(u = x^3\) with the matching differential substitution \(du = 3x^2 dx\text{.}\) Since only \(x^2 dx \) shows up in the integral, I divide by \(3\) to get \(\frac{1}{3} du = x^2 dx\text{.}\) I also need to change the bounds. When \(x = -1\text{,}\) \(u = (-1)^3 + 1 = 0\text{.}\) When \(x = 2\text{,}\) \(u = (2)^3 + 1 = 9\text{.}\) All this information lets me perform the substitution.

\begin{equation*} \int_{-1}^2 e^{x^3+1} (x^2dx) = \int_0^9 e^u \frac{1}{3} du \end{equation*}

Then I can factor out the constant and use the fact that \(e^u\) is its own antiderivatives. Finally, I evaluate the antiderivatives on the new bounds.

\begin{equation*} = \left. \frac{e^u}{3} \right|_0^9 = \frac{e^9}{3} - \frac{1}{3} = \frac{e^9-1}{3} \end{equation*}

Example 3.2.7.

\begin{equation*} \int_0^{\frac{\pi}{4}} \frac{\sin x}{\cos^3 x} dx \end{equation*}

I use the substituton \(u = \cos x\) with matching differential substitution \(du = -\sin x dx \text{.}\) Since there isn’t a negative in the the integral, I can adjust this by multiplying by \(-1\) to get \(-du = \sin x dx \text{.}\) I also need to change the bounds. When \(x = 0\text{,}\) \(u = \cos 0 = 1\text{,}\) and when \(x = \frac{\pi}{4}, u = \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}\text{.}\) All this lets me perform the substitution.

\begin{equation*} \int_0^{\frac{\pi}{4}} \frac{\sin x}{\cos^3 x} dx = \int_1^{\frac{1}{\sqrt{2}}} \frac{-1}{u^3} du \end{equation*}

Then I can take the negative out and write \(u^{-3}\) to use the inverse power rule to find the antiderivative. Finally, I evaluate the antiderivative on the new bounds.

\begin{equation*} = \left. \frac{1}{2u^2} \right|_1^{\frac{1}{\sqrt{2}}} = \frac{1}{2\frac{1}{2}} - \frac{1}{2} = \frac{1}{1} - \frac{1}{2} = \frac{1}{2} \end{equation*}

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