Chapter 5.1, Problem 5P

${\displaystyle \int \frac{K\text{dr}}{\sqrt{1-k^2{r}^2}}={\sinh }^{-1}Kr{\text{ or -cos}}^{-1}Kr{\text{ or tan}}^{-1}}\frac{K\text{r}}{\sqrt{1-k^2{r}^2}}$

Hints: Sketch a right triangle with acute angles u and v and label the sides so that $\sin u=Kr$.

Also note that $u+v=\pi /2$; then if u is an indefinite integral, so is −v since they differ by a constant of integration.