{Use of Tech} Using the integral of sec³u By reduction formula 4 in Section 8.3,
∫sec³u du = 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.
f(x) = 1/(x√(x² - 36)), [12/√3 , 12]

