Chapter 7.4, Problem 73E

Using the integral of sec³ u By reduction formula 4 in Section 7.3,

${\displaystyle \int {\sec }^{3}udu=\frac{1}{2}(\sec u\tan u+\ln |\sec u+\tan u|)+C.}$

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

73. $f(x)={(9-x^2)}^{-2},[0,\frac{3}{2}]$