`r(theta) = sec(theta)tan(theta) - 2e^theta` Find the most general antiderivative of the function. (Check your answer by differentiation.)

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Chapter 4, 4.9 - Problem 16 - 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 `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.`

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