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**flatulant guest****Guest**

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?

**irspow****Member**- Registered: 2005-11-24
- Posts: 456

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.

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**fatulant****Member**- Registered: 2005-12-01
- Posts: 10

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

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

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.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

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 (2006-02-11 07:07:40)*

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

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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