I ln(secx+ tgx)dx= tgx*ln(secx+tgx) -1/cosx
On 7/31/07, saulo nilson <[EMAIL PROTECTED]> wrote: > > w=lnsecx+tgx > dw= 1/secx+tgx * (-1/cosx^2 *-senx +secx^2)dx= > dw= 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 *coshw dw > u= w > du=dw > dv=coshwdw > v= senhw > I ln(sec x+tgx)dx= w*senhw-Isenhwdw= wsenhw-coshw > =`(n(secx+tgx) -1/cosx > > > On 7/31/07, antonio ricardo < [EMAIL PROTECTED]> > wrote: > > > > ola > > poderiam me ajudar a resolver a seguinte integral > > > > integral de ln(secx + tgx) > > > > valeu > > > > Alertas do Yahoo! Mail em seu celular. Saiba > > mais<http://br.mobile.yahoo.com/mailalertas/>. > > > > > >
