Calculus: Early Transcendentals, Chapter 4, 4.9, Section 4.9, Problem 16

The most general antiderivative R(theta) of the function r(theta) can be found using the following relation:
int r(theta)d theta = R(theta) + c
int (sec theta*tan theta - 2 e^theta)d theta = int (sec theta*tan theta)d theta - int (2 e^theta)d theta
You need to use the following formulas:
sec theta = 1/(cos theta)
tan theta = (sin theta)/(cos theta)
sec theta*tan theta = (sin theta)/(cos^2 theta)
int (sec theta*tan theta)d theta = int(sin theta)/(cos^2 theta) d theta
You need to solve the indefinite integral int(sin theta)/(cos^2 theta) d theta by substitution cos theta = t => -sin theta d theta = dt . Replacing the variable yields:
int(sin theta)/(cos^2 theta) d theta= int (-dt)/(t^2) = 1/t + c = 1/(cos theta) + c
int (2 e^theta)d theta = 2e^theta + c
Gathering all the results yields:
int (sec theta*tan theta - 2 e^theta)d theta = 1/(cos theta) -2e^theta + c
Hence, evaluating the most general antiderivative of the function yields R(theta) = 1/(cos theta) -2e^theta + c.