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

- jamiebookeater

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- anonymous

\[\int\limits \tan x dx = \int\limits \frac{ \sin x }{ \cos x } dx\]

- 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

i know i even get the answer \[\ln (\sec) + C\]
But I can't find any error

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## More answers

- anonymous

read the question it says whts the error in converting tan to sin/cos

- 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

we need calculus teachers here...not the correct answer

- P0sitr0n

no brackets should be integral (sinx/cosx)dx

- anonymous

what

- P0sitr0n

\[\int\limits tanx dx = \int\limits (\frac{ sinx }{ cosx } )dx\]

- anonymous

cmon this is not the answer this doesn't even look like an error dude

- P0sitr0n

lol just guessing maybe its this

- anonymous

dude i m frikin out lol

- 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

correction \[-ln|cosx|+C\]

- anonymous

ya so whats the error

- 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

i know i know...i got this answer too...but i can't understand what is the error...read the question

- 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

it asks for the error in converting tan to sin/cos

- 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

ok i guess...i said the same thing to my teacher but she said no there is one

- anonymous

haha

- FibonacciChick666

ha, really? that is odd. did you google it?

- anonymous

ya no help from there...google told me to come here haha

- 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

what lol...say it again confused

- FibonacciChick666

h/o a minute

- anonymous

ok ok

- 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

lol...no idea to this man.....

- anonymous

haha...i am never gonna figure it out XD

- anonymous

I will copy it and show it to my lecturer and will get back to you

- anonymous

sweet...dont forget to get the equation from comment box

- anonymous

yeah, I copied it

- anonymous

thx tell me ASAP plz

- anonymous

@amistre64 can u solve this

- 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

.... that i can determine

- anonymous

haha...but there is an error idk wht

- 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

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

@inkyvoyd ...that would make so much more sense

- 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

did you ask your instructor?

- inkyvoyd

The error is that "an error exists" itself then, I believe. Please ask your instructor :)

- 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

just watch this question

- anonymous

There is no error in the OP. int(tanx)=int(sinx/cosx)=lnabs(secx)

- anonymous

maby your getting -ln(abs(cosx)) but using the properties of logarithms this equals ln(abs(cos^(-1)x)) = ln(abs(secx))

- anonymous

wrong answer

- anonymous

will reply the correct one asap

- 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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