Chapter 8.4, Problem 74E

Using the integral of sec³ u By reduction formula 4 in Section 8.3, ${\displaystyle \int {\sec }^{3}u}du\text{ = }\frac{1}{2}\left(\sec u\text{ tan }u\text{ +ln}\left|\sec u+\tan u\right|\right)+C$.

Graph the following functions and find the area under the curve on the given interval.

74. $f\left(x\right)=\frac{1}{x\sqrt{x^2-36}}$, $\left[\frac{12}{\sqrt{3}},12\right]$