How do you find the length of the curve #y=lnabs(secx)# from #0<=x<=pi/4#?

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Answer 1 · 2017-02-02T12:39:42

#ln(sqrt2+1)=0.8814# length units, nearly..

In the given interval,# |secx|=secx#.

#y'=1/secx (secx)'=secxtanx/secx=tanx#1

Length = #int sqrt(1+(y')^2) dx#,

with #y = ln|secx| # and x from #0 to pi/4#

#=int sqrt(1+tan^2x) dx#, for the limits

#=int secx dx#, for the limita

#=[ln(secx+tanx)],# between #x = 0 and pi/4#

#=ln(sec(pi/4)+tan(pi/4))=ln(sqrt2+1)#