Single Variable Calculus with Early Transcendentals

Solutions Manual

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Sample solution:

5.4: Indefinite Integrals and the Substitution Rule – Exercise #27

If we let $u={\displaystyle \frac{1}{x}}$, then $\mathit{d}u=-{\displaystyle \frac{1}{x^2}}\mathit{d}x$ and $-\mathit{d}u={\displaystyle \frac{1}{x^2}}\mathit{d}x$. We evaluate the intergral as follows.

$\begin{array}{l}{\displaystyle \int {\displaystyle \frac{1}{x^2}}{\sec }^2\left({\displaystyle \frac{1}{x}}\right)\mathit{d}x}={\displaystyle \int {\sec }^2\left({\displaystyle \frac{1}{x}}\right)\cdot {\displaystyle \frac{1}{x^2}\mathit{d}x}}={\displaystyle \int {\sec }^{2}u\cdot -\mathit{d}u}=-{\displaystyle \int {\sec }^{2}u\mathit{d}u}\\ \phantom{{\displaystyle \int {\displaystyle \frac{1}{x^2}}{\sec }^2\left({\displaystyle \frac{1}{x}}\right)\mathit{d}x}}=-\tan u+C=-\tan \left({\displaystyle \frac{1}{x}}\right)+C\end{array}$