\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}