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\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}