Module 3: Introduction

The goal of this module is to review the integration by substitution technique, also referred to as u substitution.

In this module you will complete the following activities:

Review the integration by substitution technique.
Practice the covered concepts through some exercises, called "Now You Try".
Complete a quiz on this topic at the end of the module.

When faced with an indefinite integral it is good practice to first try Integration by substitution. We use the technique to make the problem at hand look like something we already know how to integrate.
The basic integrals we assume that you know are:

$\begin{aligned} \int u^{n} d u = \frac{1}{n + 1} u^{n + 1} + C, n \neq - 1 \\ \int u^{- 1} d u = \ln (| u |) + C \\ \int e^{u} d u = e^{u} + C \\ \int a^{u} d u = \frac{a^{u}}{\ln a} + C; (don't memorize: use a^{u} = e^{u \ln a}) \\ \int \sin (u) d u = - \cos (u) + C \\ \int \cos (u) d u = \sin (u) + C \\ \int \tan (u) d u = - \ln | c o s (u) | + C \\ \int \sec^{2} (u) = \tan (u) + C \\ \int \frac{1}{1 + u^{2}} d u = \arctan (u) + C \\ \int \frac{1}{\sqrt{1 - u^{2}}} d u = \arcsin (u) + C \end{aligned}$ $\begin{align*} &\int u^n\; du =\frac{1}{n+1}u^{n+1}+C,\;\; n \neq -1&\\ &\int u^{-1}\; du= \ln (|u|) +C\\ &\int e^u\; du =e^u+C\\ &\int a^u\; du=\frac{a^u}{\ln a}+C; \hspace{2cm} \mbox{(don't memorize: use}\; \; a^u=e^{u\ln a})\\ &\int \sin(u)\; du= -\cos(u) +C\\ &\int \cos(u)\; du=\sin(u)+C\\ &\int \tan(u)\; du=-\ln|cos(u)|+C\\ &\int \sec^2(u)=\tan(u)+C\\ &\int \frac{1}{1+u^2} \; du=\arctan(u)+C\\ &\int \frac{1}{\sqrt{1-u^2}} \; du=\arcsin(u)+C \end{align*}$