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An error exists in the logic that says the following. Explain where and why the error occurs and provide graphical support for your explanation.

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\[\int\limits \tan x dx = \int\limits \frac{ \sin x }{ \cos x } dx\]
Set u =cos x, then du = -sinx dx, so \(\int \dfrac{\sin x}{\cos x}dx=-\int \dfrac{-\sin xdx}{\cos x}=-\int \dfrac{du}{u}\). Do you recognise what to do now?
i know i even get the answer \[\ln (\sec) + C\] But I can't find any error

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Other answers:

read the question it says whts the error in converting tan to sin/cos
Well, if I go on with what I started, I get -ln|u| +C. If u > 0, this becomes \(-\ln u + C = \ln u^{-1} + C = \ln \frac{1}{u}+C=\ln(\sec x) + C\). If u < 0, you have \(-\ln(-u)+C=...=\ln(-\sec x)+C\) So it is necessary to take into cosideration what sign tan x has.
we need calculus teachers here...not the correct answer
no brackets should be integral (sinx/cosx)dx
\[\int\limits tanx dx = \int\limits (\frac{ sinx }{ cosx } )dx\]
cmon this is not the answer this doesn't even look like an error dude
lol just guessing maybe its this
dude i m frikin out lol
I am unable to find a flaw in the logic only the math, you should end up with \[ln|cosx|+C\] which you do after integration, I would assume this method approprite
correction \[-ln|cosx|+C\]
ya so whats the error
Whenever you do the u substitution integration technique it follows: let u=cosx du=-sinx we then have\[-\int\limits_{}^{}\frac{ 1 }{ u }du\] simply integrate from here the back-substitute yielding \[-ln|u|+C\] then \[-ln|cos x|+C\]
i know i know...i got this answer too...but i can't understand what is the the question
I did and there does not exist an error if it supplies a correct answer we use this trick a lot in higher exponential tangent cases-I think though i may be wrong
it asks for the error in converting tan to sin/cos
yes I understand but as I said I use this alot, I do not see the flaw in the logic provided the cosine function does not equal zero on your bounds
ok i guess...i said the same thing to my teacher but she said no there is one
ha, really? that is odd. did you google it?
ya no help from told me to come here haha
lol ok well the only issue would be the graph contains asymptotes and DNE at multiple points aka cos x=0 but your domain should be restricted to allow for this. That would be my best guess as to the answer they are fishing for
what lol...say it again confused
h/o a minute
ok ok
so if you look at the graph of the tangent function, it has horizontal asymptotes at every point where cos x=0 (shocking I know). Because cosine has a continuous domain and tangent does not, this must be accounted for whenever you evaluate a definite integral;however, your domain in a tangent function integration, usually accounts for this to start with-unless they are being...not nice people. idea to this man.....
haha...i am never gonna figure it out XD
I will copy it and show it to my lecturer and will get back to you
sweet...dont forget to get the equation from comment box
yeah, I copied it
thx tell me ASAP plz
@amistre64 can u solve this
i cannot see a reason why there would be an error in that. By definition: \[\tan x=\frac{\sin x}{\cos x}\] is a trig identity. They have the same domains so reiterating the domain as Fibonacci pointed out ... seems moot. There is no error that can determine
.... that i can determine
haha...but there is an error idk wht
@Best_Mathematician: A wide variety of Openstudy members have answered your question. The error you mention cannot be pointed out by any of them. I think it's your turn now. Just saying: there is an error is not enough. Ask your teacher what he/she means with the error. BTW: I am speaking out of experience: sometimes even teachers are wrong :) I know, because I'm a teacher myself...
are you sure it does not say int tan x dx=int sin x dx/int cos x dx? Because that would be obviously flawed.
@inkyvoyd ...that would make so much more sense
i m heck sure...the question i wrote is the exact same copy...and if it wud be like tht it wud be easy not challenging...and here i m asking challenging questions
did you ask your instructor?
The error is that "an error exists" itself then, I believe. Please ask your instructor :)
no worry... i will post the answer after my instructor replies...its spring break over here....but still i will reply the answer as soon as i can
just watch this question
There is no error in the OP. int(tanx)=int(sinx/cosx)=lnabs(secx)
maby your getting -ln(abs(cosx)) but using the properties of logarithms this equals ln(abs(cos^(-1)x)) = ln(abs(secx))
wrong answer
will reply the correct one asap
if anything, an error might exist in the assumption that an indefinite integration has a definite answer. \[\int tan(x)~dx=xxx+K\]\[\int \frac{sin(x)}{cos(x)}~dx=xxx+C\] but that distinction would be made pointless by the fact that an indefinite integral represents a family, or set, of functions anyway.

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