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Topic review (newest first)

2006-02-12 06:36:07

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.

2006-02-12 06:06:29

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.

2006-02-12 05:02:39

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.

2006-02-12 04:52:33

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

2006-02-12 01:00:19

What you are looking for can be found here:

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

flatulant guest
2006-02-11 11:43:20

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?

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