Subsection 7.4.1 Change of variables for indefinite integrals
We start by exploring some examples where we can get the desired result by the guess and check technique.
Example 7.4.1. Power of a linear by guess and check.
Find \(\int (3x+5)^7\, dx\text{.}\)
Solution.
We could do this problem by rewriting the integrand as an explicit seventh degree polynomial and then using the power and sum rules, but that is too much work. Instead, I will notice the integrand looks almost like a power, and thus guess an answer of \(\frac{1}{8} (3x+5)^8+C\text{.}\) I then check by differentiating. Using the chain rule,
\begin{equation*}
\frac{d}{dx} (\frac{1}{8} (3x+5)^8+C)=\frac{1}{8}*8(3x+5)^{8-1}*3=3(3x+5)^7\text{.}
\end{equation*}
Thus our guess was off by a factor of 3 and the correct antiderivative is
\begin{equation*}
\frac{1}{3}*\frac{1}{8} (3x+5)^8+C=\frac{1}{24} (3x+5)^8+C\text{.}
\end{equation*}
We can easily use the same trick to produce a rule for powers of a linear polynomial.
Example 7.4.2. Power of a generic linear by guess and check.
Find \(\int (ax+b)^n dx\text{.}\)
Solution.
As we did in the previous example, we first guess the antiderivative to be \(\frac{1}{n+1} (ax+b)^{n+1}+C\text{.}\) We then take the derivative of that expression and obtain \(a(ax+b)^n\text{.}\) This misses our integrand by a factor of \(a\text{.}\) We adjust by that factor and find the antiderivative is \(\frac{1}{a} \frac{1}{(n+1)} (ax+b)^{n+1}+C\text{.}\)
We can use the same trick to produce a rule for functions that are the exponential of a linear function.
Example 7.4.3. Antidifferentiation of an exponential function by guess and check.
Find \(\int e^{ax+b} dx\text{.}\)
Solution.
As we did in the last example, our first guess uses the basic rule without worrying about the linear term, so we guess \(e^{ax+b}+C\text{.}\) We then take the derivative of that expression and obtain \(ae^{ax+b}\text{.}\) This misses our integrand by a factor of \(a\text{.}\) We adjust by that factor and find the antiderivative is \(\frac{1}{a} e^{ax+b}+C\text{.}\)
We run into a problem if we try to extend this method with quadratic terms. If we start with \((x^2+5)^3\) and guess an antiderivative of \(\frac{1}{4} (x^2+5)^4\text{,}\) when we differentiate we get \((x^2+5)^3 2x\) and are off by a factor of \(8x\text{.}\) However, when we divide by that factor to get \(\frac{(x^2+5)^4}{8x}\) as a proposed antiderivative, and then differentiate again, we get
\begin{equation*}
\frac{4*2x(x^2+5)^3*8x-(x^2+5)^4*8}{8x}\text{,}
\end{equation*}
which is not what we want. The key is to start by recalling the chain rule:
\begin{equation*}
\frac{d}{dx} (f(g(x)))=f'(g(x))g'(x)\text{.}
\end{equation*}
We want to use the same rule with a different notation, using implicit differentiation and a new variable \(u\text{:}\)
\begin{equation*}
\frac{d}{dx} (f(u))=f'(u)\frac{du}{dx}\text{.}
\end{equation*}
By the fundamental theorem of calculus, we can convert this to an integration formula:
\begin{equation*}
\int f' (u) \frac{du}{dx} dx=f(u)+C\text{.}
\end{equation*}
We will generally simplify \(\frac{du}{dx} dx\) to \(du\text{,}\) so our substitution rule is
\begin{equation*}
\int f' (u)du=f(u)+C\text{.}
\end{equation*}
Let us rework some earlier examples with this method and then illustrate the method with a more difficult problem.
Example 7.4.4. Power of linear example redone with change of variables.
Find \(\int (3x+5)^7 dx\text{.}\)
Solution.
The obvious candidate for \(u\) is \(3x+5\text{.}\) Then \(du=3\,dx\text{.}\) Thus
\begin{align*}
\int (3x+5)^7\, dx \amp =\frac{1}{3} \int (3x+5)^7 (3\,dx) \quad \text{ (Make } u \text{ and } du \text{explicit.)}\\
\amp =\frac{1}{3} \int (u)^7 \, du \quad \text{ (Do the substitution.)}\\
\amp =\frac{1}{3*8} (u)^8+C \quad \text{ (Find the integral in terms of } u.)\\
\amp =\frac{1}{24} (3x+5)^8+C \quad \text{ (Substitute back.)}\text{.}
\end{align*}
This is easy to generalize for a power of a linear term.
Example 7.4.5. Power of generic linear example redone with change of variables.
Find \(\int (ax+b)^n dx\text{.}\)
Solution.
The obvious candidate for \(u\) is \(ax+b\text{.}\) Then \(du=a \, dx\text{.}\) Hence
\begin{align*}
\int (ax+b)^n\, dx \amp =\frac{1}{a} \int (ax+b)^n (a\, dx) \quad \text{ (Make } u \text{ and } du \text{ explicit.)}\\
\amp =\frac{1}{a} \int (u)^n \, du \quad \text{ (Do the substitution.)}\\
\amp =\frac{1}{a*(n+1)} (u)^{n+1}+C \quad \text{ (Find the integral in terms of } u.)\\
\amp =\frac{1}{a*(n+1)} (ax+b)^{n+1}+C \quad \text{ (Substitute back.)}\text{.}
\end{align*}
To use this method with \(u\) replacing something more complicated than a linear term, we need to have \(du\) available, with the possible addition of multiplying by a scalar constant.
Example 7.4.6. Power of cubic function with change of variables.
Find \(\int (2x^3+11)^7 x^2 dx\text{.}\)
Solution.
The obvious candidate for \(u\) is \(2x^3+11\text{,}\) since it is an expression taken to a large power. Then \(du=6x^2\, dx\text{.}\) Thus
\begin{align*}
\int (2x^3+11)^7 x^2\, dx \amp =\frac{1}{6} \int (2x^3+11)^7 (6x^2 \,dx) \quad \text{ (Make } u \text{ and } du \text{ explicit.)}\\
\amp =\frac{1}{6} \int (u)^7 \, du \quad \text{ (Do the substitution.)}\\
\amp =\frac{1}{6*8} (u)^8+C \quad \text{ (Find the integral in terms of } u.)\\
\amp =\frac{1}{48} (2x^3+11)^8+C \quad \text{ (Substitute back.)}\text{.}
\end{align*}
By convention, \(u\) is often used the new variable used with this change of variables technique, so the technique is often called \(u\)-substitution.
Subsection 7.4.2 Change of variables for definite integrals
In the definite integral, we understand that \(a\) and \(b\) are the \(x\)-values of the ends of the integral. We could be more explicit and write \(x=a\) and \(x=b\text{.}\) The last step in solving a definite integral is to substitute the endpoints back into the antiderivative we have found. We can either change the variables for the endpoints as well, or we can convert the antiderivative back to the original variables before substituting. Consider the following example.
Example 7.4.7. A definite integral with change of variables.
Evaluate \(\int_1^3 e^{2x+5} dx\text{.}\)
Solution 1.
Solution 1: Convert everything to \(u\text{.}\) The obvious candidate for \(u\) is \(2x+5\text{.}\) Then \(du=2\, dx\text{.}\) For the lower endpoint, \(x=1\) becomes \(u=2(1)+5=7\text{.}\) For the upper endpoint \(x=3\) becomes \(u=2(3)+5=11\text{.}\) Substituting,
\begin{align*}
\int_1^3 e^{2x+5}\, dx \amp =\left({\frac{1}{2}}\right) (2)\int_1^3 e^{2x+5} \,dx)\quad \text{ (Add needed factors.)}\\
\amp =\frac{1}{2} \int_1^3 e^{2x+5} (2\,dx) \quad \text{ (Make } u \text{ and } du \text{ explicit.)}\\
\amp =\frac{1}{2} \int_{u=7}^{u=11}e^u \,du \quad \text{ (Do the substitution.)}\\
\amp =\left.\frac{1}{2} e^u\right|_7^{11} \quad \text{ (Find the antiderivative.)}\\
\amp =\frac{1}{2} e^{11}-\frac{1}{2} e^7. \quad \text{ (Evaluate.)}\text{.}
\end{align*}
Solution 2.
Solution 2: Keeping, but labeling, the endpoints. We have the same \(u\) and \(du\text{,}\) but do not convert the endpoints. To reduce confusion we make sure to label the variable when we are using both \(x\) and \(u\text{.}\) Thus,
\begin{align*}
\int_1^3 e^{2x+5} \, dx \amp =\frac{1}{2} \int_1^3 e^{2x+5} (2\,dx) \quad \text{ (Make } u \text{ and } du \text{ explicit.)}\\
\amp =\frac{1}{2} \int_{x=1}^{x=3}e^u \,du \quad \text{ (Do the substitution.)}\\
\amp =\left.\frac{1}{2} e^u\right|_{x=1}^{x=3} \quad \text{ (Find the antiderivative.)}\\
\amp =\left.\frac{1}{2} e^{2x+3}\right|_{x=1}^{x=3} \quad \text{ (Convert back.)}\\
\amp =\frac{1}{2} e^{11}-\frac{1}{2} e^7. \quad \text{ (Evaluate.)}\text{.}
\end{align*}
It should be noted that when we change variables we may find ourselves looking at an integral from \(a\) to \(b\) where the \(b \lt a\text{.}\) We do not change the order of the endpoints.
Example 7.4.8. A second definite integral with change of variables.
Evaluate \(\int_{-2}^1 x e^{(x^2)}dx\)
Solution.
(Convert everything to \(u\text{.}\)) The obvious candidate for \(u\) is \(x^2\text{.}\) Then \(du=2x\, dx\text{.}\) For the lower endpoint, \(x=-2\) becomes \(u=(-2)^2=4\text{.}\) For the upper endpoint \(x=1\) becomes \(u=1^2=1\text{.}\) Substituting,
\begin{align*}
\int_{-2}^1 x e^{(x^2)}\,dx \amp =\frac{1}{2} \int_{-2}^1 e^{(x^2)}(2x\, dx) \quad \text{(Make } u \text{ and } du \text{explicit.)}\\
\amp =\frac{1}{2} \int_{4}^{1}e^u\, du \quad \text{ (Do the substitution.)}\\
\amp =\left.\frac{1}{2} e^u\right|_4^{1} \quad \text{ (Find the antiderivative.)}\\
\amp =\frac{1}{2} (1-e^4).\amp \quad \text{ (Evaluate.)}\text{.}
\end{align*}