Calculus: Early Transcendentals, Chapter 4, 4.9, Section 4.9, Problem 17

The most general antiderivative H(theta) of the function h(theta) can be found using the following relation:
int h(theta)d theta = H(theta) + c
int (2sin theta - sec^2 theta)d theta = int (2sin theta)d theta - int (sec^2 theta)d theta
You need to use the following formulas:
intsin theta d theta = -cos theta + c => int (2sin theta)d theta = -2cos theta + c
sec^2 theta = 1/(cos^2 theta) = (tan theta)' => int sec^2 theta d theta = int (tan theta)' = tan theta + c

Gathering all the results yields:
int (2sin theta - sec^2 theta)d theta =-2cos theta - tan theta + c
Hence, evaluating the most general antiderivative of the function yields H(theta) = -2cos theta - tan theta + c .