- #1

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- 12

number 13 of 8.1 says this

y= ln(secx)

find arc length from 0-pi/4

here is what i do first in my opinion.

y`= 1/sec(sectan)

y`= tanx

now i have √1+(y`)^2 = √1+tan^2

put it in an integral

∫√(1+tan^2)dx

according to books solution manual i do the following but i disagree

√1+tan^2 = √sec^2= secx

now they claim that the ∫secx = ln|secx + tanx| which i agree with. Now they say that is the answer HOWEVER, i say it is wrong because of one thing.

if the use a trig sub for back from chapter 7.3, they say you need to put a sub for dx as well which would be sec^2 so now the intergral is ∫sec^3x which is by parts...

do i not need to sub the dx because i did not start with 1+x^2 ??

thanks guys