anonymous
  • anonymous
An error exists in the logic that says the following. Explain where and why the error occurs and provide graphical support for your explanation.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits \tan x dx = \int\limits \frac{ \sin x }{ \cos x } dx\]
ZeHanz
  • ZeHanz
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?
anonymous
  • anonymous
i know i even get the answer \[\ln (\sec) + C\] But I can't find any error

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anonymous
  • anonymous
read the question it says whts the error in converting tan to sin/cos
ZeHanz
  • ZeHanz
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.
anonymous
  • anonymous
we need calculus teachers here...not the correct answer
P0sitr0n
  • P0sitr0n
no brackets should be integral (sinx/cosx)dx
anonymous
  • anonymous
what
P0sitr0n
  • P0sitr0n
\[\int\limits tanx dx = \int\limits (\frac{ sinx }{ cosx } )dx\]
anonymous
  • anonymous
cmon this is not the answer this doesn't even look like an error dude
P0sitr0n
  • P0sitr0n
lol just guessing maybe its this
anonymous
  • anonymous
dude i m frikin out lol
FibonacciChick666
  • FibonacciChick666
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
FibonacciChick666
  • FibonacciChick666
correction \[-ln|cosx|+C\]
anonymous
  • anonymous
ya so whats the error
FibonacciChick666
  • FibonacciChick666
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 \[-ln|cos x|+C\]
anonymous
  • anonymous
i know i know...i got this answer too...but i can't understand what is the error...read the question
FibonacciChick666
  • FibonacciChick666
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
anonymous
  • anonymous
it asks for the error in converting tan to sin/cos
FibonacciChick666
  • FibonacciChick666
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
anonymous
  • anonymous
ok i guess...i said the same thing to my teacher but she said no there is one
anonymous
  • anonymous
haha
FibonacciChick666
  • FibonacciChick666
ha, really? that is odd. did you google it?
anonymous
  • anonymous
ya no help from there...google told me to come here haha
FibonacciChick666
  • FibonacciChick666
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
anonymous
  • anonymous
what lol...say it again confused
FibonacciChick666
  • FibonacciChick666
h/o a minute
anonymous
  • anonymous
ok ok
FibonacciChick666
  • FibonacciChick666
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.
anonymous
  • anonymous
lol...no idea to this man.....
anonymous
  • anonymous
haha...i am never gonna figure it out XD
anonymous
  • anonymous
I will copy it and show it to my lecturer and will get back to you
anonymous
  • anonymous
sweet...dont forget to get the equation from comment box
anonymous
  • anonymous
yeah, I copied it
anonymous
  • anonymous
thx tell me ASAP plz
anonymous
  • anonymous
@amistre64 can u solve this
amistre64
  • amistre64
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
amistre64
  • amistre64
.... that i can determine
anonymous
  • anonymous
haha...but there is an error idk wht
ZeHanz
  • ZeHanz
@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
  • inkyvoyd
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.
FibonacciChick666
  • FibonacciChick666
@inkyvoyd ...that would make so much more sense
anonymous
  • anonymous
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
FibonacciChick666
  • FibonacciChick666
did you ask your instructor?
inkyvoyd
  • inkyvoyd
The error is that "an error exists" itself then, I believe. Please ask your instructor :)
anonymous
  • anonymous
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
anonymous
  • anonymous
just watch this question
anonymous
  • anonymous
There is no error in the OP. int(tanx)=int(sinx/cosx)=lnabs(secx)
anonymous
  • anonymous
maby your getting -ln(abs(cosx)) but using the properties of logarithms this equals ln(abs(cos^(-1)x)) = ln(abs(secx))
anonymous
  • anonymous
wrong answer
anonymous
  • anonymous
will reply the correct one asap
amistre64
  • amistre64
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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