`int sec(y)(tan(y) - sec(y)) dy` Find the indefinite integral.

Textbook Question

Chapter 4, 4.1 - Problem 30 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the indefinite integral, such that:

`int sec y(tan y - sec y) dy= int (sin y - 1)/(cos^2 y) dy = int (sin y)/(cos^2 y) dy - int 1/(cos^2 y) dy`

You need to solve `int (sin y)/(cos^2 y) dy` , using substitution `cos y = t => -sin ydy = dt`

`int (sin y)/(cos^2 y) dy = int (-dt)/(t^2) = 1/t + c = 1/(cos y) + c`

You need to remember that 1`/(cos^2 y) = (tan y)'`

`int sec y(tan y - sec y) dy= 1/(cos y) - tan y + c`

Hence, evaluating the indefinite integral yields `int sec y(tan y - sec y) dy= 1/(cos y) - tan y + c.`

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