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Handout Daily Prep 5.3 - Integration by Substitution

Section Overview

In this section, we see how it is possible to reverse the chain rule. This technique is called integration by substitution. Knowing how to antidifferentiate basic functions and the substitution rule together will allow us to antidifferentiate more complex functions.

Section Basic learning objectives

These are the tasks you should be able to perform with reasonable fluency when you arrive at our next class meeting. Important new vocabulary words are indicated in italics.
  • Recognize the notation for an indefinite integral and state its meaning.
  • Determine the general antiderivative given a composite function whose β€œinner” function is linear.

Section To prepare for class

Complete all actions listed below. Respond to the questions highlighted with Submit.

Checkpoint 147. Reverse the Chain Rule.

Section After class

Solidifying the concepts discussed in class through practice is necessary to build your skills.

Section Advanced learning objectives

In addition to mastering the basic objectives, here are the tasks you should be able to perform after class, with practice:
  • Use proper notation when using Integration by Substitution. In particular, fully convert an integral from \(x\)’s to \(u\)’s and back again, without mixing the two variables.
  • Use Integration by Substitution in unusual cases, such as those where there is not an obvious substitution.

Section Additional suggestions

Section Answers

Subsection To prepare for class

  1. The blue and green regions have exactly the same area.
    1. \(\displaystyle \sin(x^{3}) + C\)
    2. Yes, it matches.

Subsection After class

  1. \(\displaystyle \int x^{3} \sqrt{x^{4}+5}\ dx\) by letting \(u(x) = x^{4} + 5 = \frac{1}{6}(x^{4}+5)^{3/2}+ C\)
  2. Let \(u(x) = x+2\text{.}\) Then \(x = u-2\text{.}\) This allows one to write the integral in terms of \(u\) rather easily. Thus, \(\displaystyle \int x\sqrt{x+2}\ dx = \frac{2}{5}(x+2)^{5/2}- \frac{4}{3}(x+2)^{3/2}+C\)
  3. Use the substitution \(u=\cos x\text{.}\) Then \(\displaystyle \int \tan(x) \ dx = \int \frac{\sin x}{\cos x}\ dx = -\int \frac{1}{u}\ du = -\ln |\cos x| + C\text{.}\)
  4. \(\displaystyle \displaystyle \int \sec(x) \ dx = \ln | \sec x + \tan x | + C\)