\begin{frame} \frametitle{The Substitution Rule for Indefinite Integrals} A slightly more interesting example: \begin{exampleblock}{} \vspace{-.5ex} \begin{talign} \int x^5\sqrt{1+x^2}\, dx \end{talign} \pause We choose $u = \pause 1+x^2$. \pause Then $u' = \pause 2x$\pause, and hence \begin{talign} \int x^5\sqrt{1+x^2}\, dx \mpause[1]{&= \int x^5\sqrt{u}\, \frac{du}{2x} } \mpause{= \frac{1}{2} \int x^4\sqrt{u}\, du } \end{talign} \pause\pause\pause What now? \pause Note that \quad$x^2 = u-1$\quad and \quad$x^4 = (x^2)^2$ \pause \begin{talign} \int &x^5\sqrt{1+x^2}\, dx = \frac{1}{2} \int x^4\sqrt{u}\, du = \mpause[1]{\frac{1}{2} \int (u-1)^2\sqrt{u}\, du }\\ \mpause{&= \frac{1}{2} \int (u^2 -2u +1)\sqrt{u}\, du } \mpause{= \frac{1}{2} \int \big( u^{\frac{5}{2}} - 2u^{\frac{3}{2}} + u^{\frac{1}{2}} \big)\, du } \\ \mpause{&= \frac{1}{2} \big( \frac{2}{7}u^{\frac{7}{2}} - 2\cdot \frac{2}{5}u^{\frac{5}{2}} + \frac{2}{3}u^{\frac{3}{2}} \big) + C } \\ \mpause{&= \frac{1}{7}(1+x^2)^{\frac{7}{2}} - \frac{2}{5}(1+x^2)^{\frac{5}{2}} + \frac{1}{3}(1+x^2)^{\frac{3}{2}} + C } \end{talign} \end{exampleblock} \end{frame}