Question

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

Answer 1

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

Answer 2

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?

Answer 3

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

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

no brackets should be integral (sinx/cosx)dx

Answer 8

what

Answer 9

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

Answer 10

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

Answer 11

lol just guessing maybe its this

Answer 12

dude i m frikin out lol

Answer 13

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

Answer 14

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

Answer 15

ya so whats the error

Answer 16

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
\[-ln|cos x|+C\]

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

haha

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

what lol...say it again confused

Answer 27

h/o a minute

Answer 28

ok ok

Answer 29

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.

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

yeah, I copied it

Answer 35

thx tell me ASAP plz

Answer 36

@amistre64 can u solve this

Answer 37

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

Answer 38

.... that i can determine

Answer 39

haha...but there is an error idk wht

Answer 40

@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...

Answer 41

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.

Answer 42

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

Answer 43

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

Answer 44

did you ask your instructor?

Answer 45

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

Answer 46

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

Answer 47

just watch this question

Answer 48

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

Answer 49

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

Answer 50

wrong answer

Answer 51

will reply the correct one asap

Answer 52

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.