# `int sec(t)(sec(t) + tan(t)) dt` Find the general indefinite integral.

### Textbook Question

Chapter 5, 5.4 - Problem 16 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find the indefinite integral, hence, you need to remember that `sec t = 1/(cos t)` , such that:

`int (sec t)(sec t + tan t) dt = int (1/(cos t))(1 + sin t)/(cos t) dt`

`int (sec t)(sec t + tan t) dt = int 1/(cos^2 t) dt + int (sint)/(cos^2 t) dt`

`int (sec t)(sec t + tan t) dt = tan t + int (sint)/(cos^2 t) dt`

You need to solve for the indefinite integral `int (sint)/(cos^2 t) dt` using the substitution `cos t = u => -sint dt = du.`

`int (sint)/(cos^2 t) dt = int (-du)/(u^2) = 1/u + c`

Replacing back `cost` for `u` yields:

`int (sint)/(cos^2 t) dt = 1/(cos t) + c`

Hence, evaluating the indefinite integral, yields `int (sec t)(sec t + tan t) dt = tan t + 1/(cos t) + c = (sin t+1)/(cos t) + c`