Homework Help

Find the integral. `int sec^3x dx`

user profile pic

masterpiece11 | Student | Salutatorian

Posted October 13, 2012 at 1:13 AM via web

dislike 3 like

Find the integral.

`int sec^3x dx`

Tagged with calculus 2, math

1 Answer | Add Yours

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted October 14, 2012 at 3:21 PM (Answer #1)

dislike 1 like

You should use the following formula such that:

`int sec^n x dx = (sin x*sec^(n-1) x)/(n-1) + (n-2)/(n-1) int sec^(n-2) x dx`

Reasoning by analogy yields:

`int sec^3 x dx = (sin x*sec^2 x)/2 + 1/2 int sec x dx`

`int sec^3 x dx = (1/2)(tan x*sec x + int sec x dx)`

`int sec x dx = int 1/(cos 2(x/2))dx`

`int sec x dx = int 1/(cos^2(x/2) - sin^2(x/2)) dx`

`int sec x dx = int 1/((cos(x/2) - sin(x/2))(cos(x/2) + sin(x/2)))`

`int sec x dx = ln(cos(x/2) + sin(x/2)) - ln(cos(x/2)- sin(x/2))`

`int sec^3 x dx = (1/2)(tan x*sec x + ln(cos(x/2) + sin(x/2)) - ln(cos(x/2) - sin(x/2))) + c`

Hence, evaluating the indefinite integral yields `int sec^3 x dx = (1/2)(tan x*sec x + ln(cos(x/2) + sin(x/2)) - ln(cos(x/2) - sin(x/2))) + c.`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes