\begin{frame} \frametitle{The Substitution Rule for Indefinite Integrals} \begin{exampleblock}{} \vspace{-.5ex} \begin{talign} \int \frac{x}{\sqrt{1-4x^2}} dx \end{talign} \pause We choose $u = \pause 1-4x^2$. \pause Then $u' = \pause -8x$\pause, and hence \begin{talign} \int \frac{x}{\sqrt{1-4x^2}} dx \mpause[1]{&= \int \frac{x}{\sqrt{u}}\, \frac{du}{-8x} } \mpause{= -\frac{1}{8} \int \frac{1}{\sqrt{u}}\, du }\\ \mpause{&= -\frac{1}{8} \cdot 2\sqrt{u} + C} \mpause{= -\frac{1}{4} \sqrt{1-4x^2} + C} \end{talign} \end{exampleblock} \pause\pause\pause\pause\pause \begin{exampleblock}{} \vspace{-.5ex} \begin{talign} \int e^{5x} dx \end{talign} \pause We choose $u = \pause 5x$. \pause Then $u' = \pause 5$\pause, and hence \begin{talign} \int e^{5x} dx \mpause[1]{&= \int e^u\, \frac{du}{5} } \mpause{= \frac{1}{5} \int e^u\, du }\\ \mpause{&= \frac{1}{5} e^u + C} \mpause{= \frac{1}{5} e^{5x} + C} \end{talign} \end{exampleblock} \end{frame}