- #1
LT72884
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Ello every one, i have interesting question. Any one who has james stewert 7th edition calc book I am on secotion 8.1 studying for an exam.
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
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