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•  » How do you get the secant and cosecant intergration formulas?

## #1 2006-02-11 11:43:20

flatulant guest
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### How do you get the secant and cosecant intergration formulas?

I do not understand how the integral of secant x is ln(secant x + tangent x)+C and how the integral of cosecant x is -ln(cosecant x + cotangent x).  Can someone please explain to me how these are derived?

## #2 2006-02-12 01:00:19

irspow
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### Re: How do you get the secant and cosecant intergration formulas?

What you are looking for can be found here:

www.math2.org/math/integrals/more/sec.htm

They say it isn't intuitive...and it isn't, but it is the proof.

## #3 2006-02-12 04:52:33

fatulant
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### Re: How do you get the secant and cosecant intergration formulas?

why multiply by sec x + tan x?  Why not sin x + cos x?  Or csc x + cot x?

## #4 2006-02-12 05:02:39

irspow
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### Re: How do you get the secant and cosecant intergration formulas?

They did say it isn't intuitive...I really don't know.  They might have chosen that identity simply for ease of simplification after integration.  I am in no way a proofs guy.

## #5 2006-02-12 06:06:29

mikau
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### Re: How do you get the secant and cosecant intergration formulas?

Basicly we begin with sec x. We multiply above and below by sec x + tan x to get:

( sec^2 x + sec x tan x  )/ (sec x + tan x )

if we let u = sec x + tan x, we can differentiate to find du = sec^2 x + sec x tan x   which happens to be the expression in the numerator, therefore we make the substitutions to get:

1/u du

this is the derivative of ln u. So the answer is ln|u|. We declared u to be sec x + tan x therefore the answer is ln|sec x + tan x|  and of course, + C.

Basicly muliplying above and below by sec x + tan x is a nifty trick that gives it the form 1/u du which is easy to integrate.

Last edited by mikau (2006-02-12 06:07:40)

A logarithm is just a misspelled algorithm.

## #6 2006-02-12 06:36:07

irspow
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### Re: How do you get the secant and cosecant intergration formulas?

Mikau is correct.  Besides if you just integrate sec x = 1 / cos x

∫1 / cosx  dx = ln[cos(x/2) + sin(x/2)] - ln[cos(x/2) - sin(x/2)] + C

It is much neater to use their solution.

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