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anonymous
 3 years ago
An error exists in the logic that says the following. Explain where and why the error occurs and provide graphical support for your explanation.
anonymous
 3 years ago
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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \tan x dx = \int\limits \frac{ \sin x }{ \cos x } dx\]

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.0Set 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i know i even get the answer \[\ln (\sec) + C\] But I can't find any error

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0read the question it says whts the error in converting tan to sin/cos

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.0Well, if I go on with what I started, I get lnu +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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we need calculus teachers here...not the correct answer

P0sitr0n
 3 years ago
Best ResponseYou've already chosen the best response.0no brackets should be integral (sinx/cosx)dx

P0sitr0n
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits tanx dx = \int\limits (\frac{ sinx }{ cosx } )dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0cmon this is not the answer this doesn't even look like an error dude

P0sitr0n
 3 years ago
Best ResponseYou've already chosen the best response.0lol just guessing maybe its this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dude i m frikin out lol

FibonacciChick666
 3 years ago
Best ResponseYou've already chosen the best response.1I am unable to find a flaw in the logic only the math, you should end up with \[lncosx+C\] which you do after integration, I would assume this method approprite

FibonacciChick666
 3 years ago
Best ResponseYou've already chosen the best response.1correction \[lncosx+C\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ya so whats the error

FibonacciChick666
 3 years ago
Best ResponseYou've already chosen the best response.1Whenever 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 backsubstitute yielding \[lnu+C\] then \[lncos x+C\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i know i know...i got this answer too...but i can't understand what is the error...read the question

FibonacciChick666
 3 years ago
Best ResponseYou've already chosen the best response.1I did and there does not exist an error if it supplies a correct answer we use this trick a lot in higher exponential tangent casesI think though i may be wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it asks for the error in converting tan to sin/cos

FibonacciChick666
 3 years ago
Best ResponseYou've already chosen the best response.1yes 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok i guess...i said the same thing to my teacher but she said no there is one

FibonacciChick666
 3 years ago
Best ResponseYou've already chosen the best response.1ha, really? that is odd. did you google it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ya no help from there...google told me to come here haha

FibonacciChick666
 3 years ago
Best ResponseYou've already chosen the best response.1lol 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what lol...say it again confused

FibonacciChick666
 3 years ago
Best ResponseYou've already chosen the best response.1so 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 withunless they are being...not nice people.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol...no idea to this man.....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha...i am never gonna figure it out XD

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I will copy it and show it to my lecturer and will get back to you

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sweet...dont forget to get the equation from comment box

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@amistre64 can u solve this

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0i 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0.... that i can determine

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha...but there is an error idk wht

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.0@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...

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0are you sure it does not say int tan x dx=int sin x dx/int cos x dx? Because that would be obviously flawed.

FibonacciChick666
 3 years ago
Best ResponseYou've already chosen the best response.1@inkyvoyd ...that would make so much more sense

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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

FibonacciChick666
 3 years ago
Best ResponseYou've already chosen the best response.1did you ask your instructor?

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0The error is that "an error exists" itself then, I believe. Please ask your instructor :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just watch this question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There is no error in the OP. int(tanx)=int(sinx/cosx)=lnabs(secx)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0maby your getting ln(abs(cosx)) but using the properties of logarithms this equals ln(abs(cos^(1)x)) = ln(abs(secx))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0will reply the correct one asap

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0if 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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