\begin{frame} \frametitle{The Substitution Rule for Indefinite Integrals} \begin{exampleblock}{} \vspace{-.5ex} \begin{talign} \int \tan x\, dx \end{talign}\vspace{-1ex}% \pause First, we note that: \begin{talign} \int \tan x\, dx = \int \frac{\sin x}{\cos x}\, dx \end{talign} \pause We choose $u = \pause \cos x$. \pause Then $u' = \pause -\sin x$\pause, and hence \begin{talign} \int \frac{\sin x}{\cos x} dx \mpause[1]{&= \int \frac{\sin x}{u}\, \frac{du}{-\sin x} } \mpause{= -\int \frac{1}{u}\, du }\\ \mpause{&= -\ln |u| + C} \mpause{= - \ln |\cos x| + C} \end{talign} \end{exampleblock} \pause\pause\pause\pause\pause Note that\vspace{-.5ex} \begin{talign} - \ln |\cos x| = \ln |\cos x|^{-1} = \ln |\sec x| \end{talign} \pause Thus we can also write: \begin{block}{} \begin{malign} \int \tan x\, dx = \ln |\sec x| + C \end{malign} \end{block} \end{frame}