`h(theta) = 2sin(theta) - sec^2 (theta)` Find the most general antiderivative of the function.

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Chapter 4, 4.9 - Problem 17 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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` .

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