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Best_Mathematician
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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.
 one year ago
 one year ago
Best_Mathematician Group Title
An error exists in the logic that says the following. Explain where and why the error occurs and provide graphical support for your explanation.
 one year ago
 one year ago

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Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits \tan x dx = \int\limits \frac{ \sin x }{ \cos x } dx\]
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
i know i even get the answer \[\ln (\sec) + C\] But I can't find any error
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
read the question it says whts the error in converting tan to sin/cos
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.0
Well, 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.
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
we need calculus teachers here...not the correct answer
 one year ago

P0sitr0n Group TitleBest ResponseYou've already chosen the best response.0
no brackets should be integral (sinx/cosx)dx
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
what
 one year ago

P0sitr0n Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits tanx dx = \int\limits (\frac{ sinx }{ cosx } )dx\]
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
cmon this is not the answer this doesn't even look like an error dude
 one year ago

P0sitr0n Group TitleBest ResponseYou've already chosen the best response.0
lol just guessing maybe its this
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
dude i m frikin out lol
 one year ago

FibonacciChick666 Group TitleBest ResponseYou've already chosen the best response.1
I 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
 one year ago

FibonacciChick666 Group TitleBest ResponseYou've already chosen the best response.1
correction \[lncosx+C\]
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
ya so whats the error
 one year ago

FibonacciChick666 Group TitleBest ResponseYou've already chosen the best response.1
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 backsubstitute yielding \[lnu+C\] then \[lncos x+C\]
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
i know i know...i got this answer too...but i can't understand what is the error...read the question
 one year ago

FibonacciChick666 Group TitleBest ResponseYou've already chosen the best response.1
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 casesI think though i may be wrong
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
it asks for the error in converting tan to sin/cos
 one year ago

FibonacciChick666 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
ok i guess...i said the same thing to my teacher but she said no there is one
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
haha
 one year ago

FibonacciChick666 Group TitleBest ResponseYou've already chosen the best response.1
ha, really? that is odd. did you google it?
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
ya no help from there...google told me to come here haha
 one year ago

FibonacciChick666 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
what lol...say it again confused
 one year ago

FibonacciChick666 Group TitleBest ResponseYou've already chosen the best response.1
h/o a minute
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
ok ok
 one year ago

FibonacciChick666 Group TitleBest ResponseYou've already chosen the best response.1
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 withunless they are being...not nice people.
 one year ago

Brotherman Group TitleBest ResponseYou've already chosen the best response.0
lol...no idea to this man.....
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
haha...i am never gonna figure it out XD
 one year ago

Brotherman Group TitleBest ResponseYou've already chosen the best response.0
I will copy it and show it to my lecturer and will get back to you
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
sweet...dont forget to get the equation from comment box
 one year ago

Brotherman Group TitleBest ResponseYou've already chosen the best response.0
yeah, I copied it
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
thx tell me ASAP plz
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
@amistre64 can u solve this
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
.... that i can determine
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
haha...but there is an error idk wht
 one year ago

ZeHanz Group TitleBest 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...
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

FibonacciChick666 Group TitleBest ResponseYou've already chosen the best response.1
@inkyvoyd ...that would make so much more sense
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

FibonacciChick666 Group TitleBest ResponseYou've already chosen the best response.1
did you ask your instructor?
 one year ago

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

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
just watch this question
 one year ago

Edutopia Group TitleBest ResponseYou've already chosen the best response.0
There is no error in the OP. int(tanx)=int(sinx/cosx)=lnabs(secx)
 one year ago

Edutopia Group TitleBest ResponseYou've already chosen the best response.0
maby your getting ln(abs(cosx)) but using the properties of logarithms this equals ln(abs(cos^(1)x)) = ln(abs(secx))
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
wrong answer
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
will reply the correct one asap
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago
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