8.1: Integration by Substitution
This activity is intended to supplement Calculus, Chapter 7, Lesson 1.
ID: 9889
Time required: 45 minutes
Topic: Techniques of Integration

Use substitutions such as \begin{align*}u = \sqrt{ax + b}\end{align*}
u=ax+b−−−−−√ to compute an integral.  Use Integral (on the Calculus menu) to verify manual calculation of integrals.
Activity Overview
In this activity, we explore methods for computing integrals of functions not in one of the standard forms. The focus here is upon the use of substitution to transform the given integral into a standard form.
Teacher Preparation
This investigation offers opportunities for review and consolidation of key concepts related to methods of substitution and integration of composite functions. Opportunities are provided for skill development and practice of the method of taking integrals of suitable functions. As such, care should be taken to provide ample time for ALL students to engage actively with the requirements of the task, allowing some who may have missed aspect of earlier work the opportunity to build new and deeper understanding.
 This activity can serve to consolidate earlier work on the product rule and methods of integration. It offers a suitable introduction to integration by substitution.
 Begin by reviewing the method of differentiation of composite functions (the “chain rule”) and the methods of integration of the standard function forms.
 This activity requires the use of CAS technology for students to check their answers.
Classroom Management
 This activity is designed to be teacherled. Have the students work in pairs. Use the following pages to present the material to the class and encourage discussion. Although the majority of the ideas and concepts are presented in this document, be sure to cover all the material necessary for students’ total comprehension.
 The students will need to be able to compute the derivative and the indefinite integral on their own.
 Answers should be completed on the student worksheet.

Before starting this activity, students should go to the home screen and select \begin{align*}F6\end{align*}
F6 :Clean Up > 2:NewProb, then press ENTER. This will clear any stored variables, turn off any functions and plots, and clear the drawing and home screens.
Associated Materials
 Student Worksheet: Integration by Substitution http://www.ck12.org/flexr/chapter/9732
Problem1 – Introduction
Step 1: Begin with discussion and review of both the chain rule for differentiation of composite functions and of the integrals of standard function form.
Ensure that students are comfortable with these and then challenge them to consider more difficult forms – in this case, composite functions of the form \begin{align*}y = f(g(x))\end{align*}
Step 2: Students are to use substitution to integrate \begin{align*}\int \limits \sqrt{2x + 3} \ dx\end{align*}
Students should use the selection of \begin{align*}u\end{align*}
Note: Some students may not realize that they need to have the \begin{align*}\frac{1}{2}\end{align*}
In addition, they should use the integral command (Home > \begin{align*}F3\end{align*}
Step 3: The integral for \begin{align*}\sin (x)\cos (x) dx\end{align*}
The third way students find the integral is by transforming \begin{align*}\sin (x)\cos (x)dx\end{align*}
Students can check the answer with their calculator.
Problem2 – Common Feature
Step 4: Students are to work through three problems using substitution to integrate. First, have them discuss what they should use for \begin{align*}u\end{align*}
Step 5: After working through the problems, students are challenged to identify the common feature: that each of the given functions in some way includes the derivative of the function substituted. It is critical for students to understand that this method will not work for all functions, but only for certain well chosen forms.
Extension
The challenge to the students in these problems is to use trig identities to rewrite the integral so that they have a substitution format. \begin{align*}\tan(x) & = \frac{ \sin(x)}{ \cos(x)} \\
\cos (x)^3 & = \cos (x) \cos (x)^2 = \cos (x)(1  \sin (x)^2)\end{align*}
Solutions

\begin{align*}u = 2x+3; du = 2 dx; \int \limits \sqrt{2x + 3} \ dx = \frac{1}{2} \int \limits u^{\frac{1}{2}} du = \frac{1}{3}u^{\frac{3}{2}} + C = \frac{1}{3}(2x + 3)^{\frac{3}{2}} + C\end{align*}
u=2x+3;du=2dx;∫2x+3−−−−−√ dx=12∫u12du=13u32+C=13(2x+3)32+C 
\begin{align*}u = \sin(x); du = \cos (x) dx; \int \limits \sin (x) \cos (x)dx = \int \limits u \ du = \frac{1}{2}u^2 + C = \frac{1}{2} \sin (x)^2 + C\end{align*}
u=sin(x);du=cos(x)dx;∫sin(x)cos(x)dx=∫u du=12u2+C=12sin(x)2+C 
\begin{align*}u = \cos(x); du =  \sin (x) dx; \int \limits \sin (x) \cos (x) dx =  \int \limits u \ du =  \frac{1}{2}u^2 + C =  \frac{1}{2} \cos (x)^2 + C\end{align*}
u=cos(x);du=−sin(x)dx;∫sin(x)cos(x)dx=−∫u du=−12u2+C=−12cos(x)2+C 
\begin{align*}u = 2x; du = 2 dx; \int \limits \frac{1}{2} \sin (2x) dx = \frac{1}{4} \int \limits \sin(u) du =  \frac{1}{4} \cos (u) + C =  \frac{1}{4} \cos (2x) + C\end{align*}
u=2x;du=2dx;∫12sin(2x)dx=14∫sin(u)du=−14cos(u)+C=−14cos(2x)+C 
\begin{align*}u = x^2 + 2x + 3; du & = (2x + 2) dx;\\
\int \limits \frac{x +1}{x^2 + 2x +3} dx & = \frac{1}{2} \int \limits \frac{1}{u} \ du = \frac{1}{2} \ln (u) + C = \frac{1}{2}\ln(x^2 + 2x + 3) + C \end{align*}
u=x2+2x+3;du∫x+1x2+2x+3dx=(2x+2)dx;=12∫1u du=12ln(u)+C=12ln(x2+2x+3)+C 
\begin{align*}u = \cos(x); du =  \sin(x)dx; \int \limits \sin(x)e^{ \cos(x)} dx =  \int \limits e^u du = e^u + C = e^{ \cos (x)} + C\end{align*}
u=cos(x);du=−sin(x)dx;∫sin(x)ecos(x)dx=−∫eudu=−eu+C=−ecos(x)+C 
\begin{align*}u = 4x^2 + 1; du = 8x \ dx; \int \limits \frac{x}{4x^2 + 1}dx = \frac{1}{8} \int \limits \frac{1}{u} du = \frac{1}{8} \ln (u) + C = \frac{1}{8}\ln(4x^2 + 1) + C \end{align*}
u=4x2+1;du=8x dx;∫x4x2+1dx=18∫1udu=18ln(u)+C=18ln(4x2+1)+C  Each contains the derivative of the substitution element: if \begin{align*}u\end{align*}
u is the substitution, then \begin{align*}\frac{du}{dx}\end{align*}dudx exists as part of the expression, or at least a constant multiple of it. 
\begin{align*}u = \cos (x); du =  \sin(x); \int \limits \frac{ \sin(x)}{ \cos(x)} dx =  \int \limits \frac{1}{u} du =  \ln(u) + C =  \ln( \cos(x)) + C\end{align*}
u=cos(x);du=−sin(x);∫sin(x)cos(x)dx=−∫1udu=−ln(u)+C=−ln(cos(x))+C  \begin{align*}u = \sin(x); du = \cos(x)dx;\end{align*}
\begin{align*}\int \limits \cos(x)(1 \sin(x)^2)dx = \int \limits(1u^2)du = u  \frac{u^3}{3} + C = \sin(x)  \frac{ \sin(x)^3}{3} + C\end{align*}
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