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Re: [obm-l] sec^n x
- To: obm-l@xxxxxxxxxxxxxx
- Subject: Re: [obm-l] sec^n x
- From: "Marcelo Salhab Brogliato" <msbrogli@xxxxxxxxx>
- Date: Wed, 20 Jun 2007 20:45:10 -0300
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Olá,
lembre que (tgx)' = sec^2x...
(sec^2x)' = (1/cos^2x)' = 2cosxsenx/cos^4x = 2tgx*sec^2x
int(sec^2x * sec^(n-2)x) = tgx * sec^(n-2)x - int(tgx * (n-2)
sec^(n-3)x * 2tgx*sec^2x)
agora, basta usar que tg^2x + 1 = sec^2x
abracos,
Salhab
On 6/19/07, Klaus Ferraz <klausferraz@yahoo.com.br> wrote:
>
> Alguem sabe deduzir a expressao recursiva da int(sec^n x). To fazendo várias
> integrações por parte mas sempre chega em alguma mais complicada q antes.
> grato.
> ________________________________
> Novo Yahoo! Cadê? - Experimente uma nova busca.
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