Re: [obm-l] integral

Subject: Re: [obm-l] integral

From: "saulo nilson" <saulo.nilson@xxxxxxxxx>

Date: Wed, 1 Aug 2007 12:01:56 -0300

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In-Reply-To: <F481C0D13C5B2340A09C98A4DBFCBC33C20465@MAIL.mme.gov.br>

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Achei o erro

w=lnsecx+tgx

dw= 1/secx+tgx * (-1/cosx^2 *-senx +secx^2)dx=

*** dw= 1/cosxdx

cosxe^w-1=rq(1-cosx^2)

e^2wcosx^2-2cosxe^w+1=1-cosx^2

cosx^2(e^2w+1)-2cosxe^w=0

cosx= 2e^w/(e^2w+1)=2/(e^w+e^-w)=1/coshw

I w *(1/coshw )dw

u= w

du=dw

dv=1/coshwdw

v= x

I ln(sec x+tgx)dx= w*x -I x*dw= w*x-I x/cosxdx

On 7/31/07, Artur Costa Steiner <artur.steiner@mme.gov.br> wrote:

Acho que tem um erro aqui. A derivada do segundo membro eh (tgx*ln(secx+tgx) -1/cosx)' = (secx)^2* ln(sec x + tan x) + tg x sec x - tg x sec x = (secx)^2* ln(sec x + tan x). Diferente, portanto, do integrando

Artur

[Artur Costa Steiner]

----Mensagem original-----
De: owner-obm-l@mat.puc-rio.br [mailto:owner-obm-l@mat.puc-rio.br]Em nome de saulo nilson
Enviada em: terça-feira, 31 de julho de 2007 14:36
Para: obm-l@mat.puc-rio.br
Assunto: Re: [obm-l] integral

I ln(secx+ tgx)dx= tgx*ln(secx+tgx) -1/cosx