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Math 220-2: Kursobjekt

Aaron Greicius

Section 1.7 Substitution

In the spirit of the De Morgan quote in Dictum 1.5.3, running various derivative rules in reverse yields different integration techniques. The substitution theorem is nothing more than the chain rule expressed in terms of antiderivatives.

Proof.

Assume \(F\) is an antiderivative of \(f\text{,}\) so that \(F'(x)=f(x)\) for all \(x\) in the domain of \(f\text{.}\) It follows that
\begin{align*} \frac{d}{dx}F(g(x)) \amp =F'(g(x))g'(x)\amp \text{(chain rule)}\\ \amp = f(g(x))g'(x) \amp (F'=f) \text{.} \end{align*}

Example 1.7.2. Substitution:straightforward.

Use the substitution technique to compute the following indefinite integrals.
  1. \(\displaystyle \displaystyle\int -\sin t\, \sqrt{\cos t}\, dt\)
  2. \(\displaystyle \displaystyle\int \frac{\sin(\sqrt{s})}{2\sqrt{s}}\, ds\)
  3. \(\displaystyle \displaystyle\int x^2\sqrt{x^3+1}\, dx\)
Solution.
  1. We have \(-\sin t\, \sqrt{\cos t}=f(g(t))g'(t)\) where \(f(t)=\sqrt{t}\) and \(g(t)=\cos t\text{.}\) Since \(F(t)=\frac{2}{3}\sqrt{x}\) is an antiderivative of \(f\text{,}\) Theorem 1.7.1 tells us that \(F(g(t))=\frac{2}{3}(\cos t)^{3/2}\) is an antiderivative of \(f(g(x))g'(x)= -\sin t\, \sqrt{\cos t}\text{.}\) We conclude
    \begin{equation*} \int -\sin t\, \sqrt{\cos t}\, dt =\frac{2}{3}(\cos t)^{3/2}+C\text{.} \end{equation*}
  2. We have \(\frac{\sin(\sqrt{s})}{2\sqrt{s}}=f(g(s))g'(s)\) where \(f(s)=\sqrt{s}\) and \(g(s)=\sqrt{s}\text{.}\) Since \(F(s)=-\cos t\) is an antiderivative of \(f\text{,}\) Theorem 1.7.1 tells us that \(F(g(s))=-\cos(\sqrt{s})\) is an antiderivative of \(f(g(x))g'(x)= \frac{\sin(\sqrt{s})}{2\sqrt{s}}\text{.}\) We conclude
    \begin{equation*} \int \frac{\sin(\sqrt{s})}{2\sqrt{s}}\, ds =-\cos(\sqrt{s})+C\text{.} \end{equation*}
  3. A similar approach as above can be applied, though here we have \(x^2\sqrt{x^3+1}=\frac{1}{3}f(g(x))g'(x)\text{,}\) where \(f(x)=\sqrt{x}\) and \(g(x)=x^3+1\text{.}\) It follows that
    \begin{equation*} \int x^2\sqrt{x^3+1}\, dx=\frac{1}{3}\int f(g(x))g'(x)\, dx=\frac{2}{9}(x^3+1)^{3/2}+C\text{,} \end{equation*}
    since \(F(x)=\frac{2}{3}x^{3/2}\) is an antiderivative of \(f(x)=\sqrt{x}\text{.}\)
The usefulness of Theorem 1.7.1 depends on your ability to express the given integrand in the very particular form \(f(g(x))g'(x)\text{.}\) This is not always as easy to do as it was in Example 1.7.2. The substitution procedure facilitates this process, giving it a more algebraic flavor.

Remark 1.7.4. Substitution procedure.

Procedure 1.7.3 is so algebraic in flavor that we sometimes forget that the chain rule is in play here! It is baked into the transformation step
\begin{equation*} \int f(x)\, dx=\int h(u)\, du\text{.} \end{equation*}
Using the substitution equations (1.7.2)–(1.7.3), we see that this equality is really just an algebraic short hand for the equality
\begin{equation*} \int f(x)\, dx= \int h(g(x))g'(x)\,dx\text{.} \end{equation*}
Now the substitution theorem tells us that if \(F\) is an antiderivative of \(h\text{,}\) then \(F(g(x))\) is an antiderivative of \(h(g(x))g'(x)\text{.}\) This is why our conclusion
\begin{equation*} \int f(x)\, dx=F(g(x))+C \end{equation*}
is valid.

Example 1.7.5. Substitution: less straightforward.

Use the substitution technique to compute the following indefinite integrals.
  1. \(\displaystyle \displaystyle\int\frac{x}{\sqrt{x+1}}\, dx\)
  2. \(\displaystyle \displaystyle\int (1+\sqrt{t})^{100}\, dt\)
Solution.
  1. We choose the substitution
    \begin{align} u \amp= x+1 \tag{1.7.4}\\ du \amp = dx \tag{1.7.5} \end{align}
    Equation (1.7.4) implies \(x=u-1\text{.}\) Transforming the integral, we get
    \begin{align*} \int\frac{x}{\sqrt{x+1}}\, dx\amp = \int \frac{u-1}{\sqrt{u}}\, du \\ \amp = \int u^{1/2}-u^{-1/2}\, du\\ \amp = \frac{2}{3}u^{3/2}-2u^{1/2}+C\\ \amp = \frac{2}{3}(x+1)^{3/2}-2(x+1)^{1/2}+C\text{.} \end{align*}
  2. We choose the substitution
    \begin{align} u \amp= 1+\sqrt{t} \tag{1.7.6}\\ du \amp = \frac{1}{2\sqrt{t}} dt \tag{1.7.7} \end{align}
    Equation (1.7.7) implies \(dt=2\sqrt{t}du\text{.}\) By (1.7.7), we have \(\sqrt{t}=u-1\text{.}\) Thus \(dt=2\sqrt{t}du=2(u-1)\text{.}\) Transforming the integral, we get
    \begin{align*} \int(1+\sqrt{t})^{100}\, dt \amp = \int u^{100}2(u-1)\, du \\ \amp = 2\int u^{101}-u^{100} du\\ \amp = \frac{2}{102}u^{102}-\frac{2}{101}u^{101}+C\\ \amp = \frac{2}{102}(1+\sqrt{t})^{102}-\frac{2}{101}(1+\sqrt{t})^{101}+C\text{.} \end{align*}
Procedure 1.7.3 is a technique for computing antiderivatives. Of course, we can combine this with FTC II to obtain a substitution method for definite integrals. In the next section we will provide an alternative, more streamlined approach to computing definite integrals using substitution.

Example 1.7.7. Definite integral substitution: 2-step technique.

Use substitution to compute \(\displaystyle\int_1^2\frac{x}{\sqrt{x+1}}\, dx\text{.}\)
Solution.
We saw in Example 1.7.5 that an antiderivative of the given integrand is
\begin{equation*} F(x)=\frac{2}{3}(x+1)^{3/2}-2(x+1)^{1/2}=2(x+1)^{1/2}\left(\frac{1}{3}x-\frac{2}{3}\right)\text{.} \end{equation*}
We conclude using FTC II that
\begin{align*} \int_1^2\frac{x}{\sqrt{x+1}}\, dx \amp =F(2)-F(1)\\ \amp = 2(3)^{1/2}\left(\frac{2}{3}-\frac{2}{3}\right)-2(2)^{1/2}\left(\frac{1}{3}-\frac{2}{3}\right)\\ \amp = \frac{2\sqrt{2}}{3}\text{.} \end{align*}
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