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\begin{frame}
  \frametitle{The Substitution Rule: Exercises}
  
  \begin{exampleblock}{}
    Evaluate
    \begin{talign}
      \int \frac{\sin 2x}{1+\cos^2 x}\, dx 
    \end{talign}
    \pause
    
    We recall that \quad $\sin 2x = 2 \sin x\cos x$
    \pause\medskip
    
    We take $u = \pause 1+(\cos x)^2$\pause.
    \medskip
    
    Then $u' = \pause 2\cos x(-\sin x)\pause = -\sin 2x$ \pause and
    \begin{talign}
      \int \frac{\sin 2x}{1+\cos^2 x}\, dx 
      \mpause[1]{ &= \int \frac{\sin 2x}{u} \, \frac{du}{-\sin 2x} }\\
      \mpause{ &= -\int \frac{1}{u} \, du }\\
      \mpause{ &= -\ln |u| + C }\\
      \mpause{ &= -\ln |1+(\cos x)^2| + C }
    \end{talign}\vspace{-2.5ex}
    \pause\pause\pause\pause%
  \end{exampleblock}
\end{frame}