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Best_Mathematician

  • 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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  1. Best_Mathematician
    • 3 years ago
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    \[\int\limits \tan x dx = \int\limits \frac{ \sin x }{ \cos x } dx\]

  2. ZeHanz
    • 3 years ago
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    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?

  3. Best_Mathematician
    • 3 years ago
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    i know i even get the answer \[\ln (\sec) + C\] But I can't find any error

  4. Best_Mathematician
    • 3 years ago
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    read the question it says whts the error in converting tan to sin/cos

  5. ZeHanz
    • 3 years ago
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    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.

  6. Best_Mathematician
    • 3 years ago
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    we need calculus teachers here...not the correct answer

  7. P0sitr0n
    • 3 years ago
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    no brackets should be integral (sinx/cosx)dx

  8. Best_Mathematician
    • 3 years ago
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    what

  9. P0sitr0n
    • 3 years ago
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    \[\int\limits tanx dx = \int\limits (\frac{ sinx }{ cosx } )dx\]

  10. Best_Mathematician
    • 3 years ago
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    cmon this is not the answer this doesn't even look like an error dude

  11. P0sitr0n
    • 3 years ago
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    lol just guessing maybe its this

  12. Best_Mathematician
    • 3 years ago
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    dude i m frikin out lol

  13. FibonacciChick666
    • 3 years ago
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    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

  14. FibonacciChick666
    • 3 years ago
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    correction \[-ln|cosx|+C\]

  15. Best_Mathematician
    • 3 years ago
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    ya so whats the error

  16. FibonacciChick666
    • 3 years ago
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    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 \[-ln|cos x|+C\]

  17. Best_Mathematician
    • 3 years ago
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    i know i know...i got this answer too...but i can't understand what is the error...read the question

  18. FibonacciChick666
    • 3 years ago
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    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

  19. Best_Mathematician
    • 3 years ago
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    it asks for the error in converting tan to sin/cos

  20. FibonacciChick666
    • 3 years ago
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    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

  21. Best_Mathematician
    • 3 years ago
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    ok i guess...i said the same thing to my teacher but she said no there is one

  22. Best_Mathematician
    • 3 years ago
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    haha

  23. FibonacciChick666
    • 3 years ago
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    ha, really? that is odd. did you google it?

  24. Best_Mathematician
    • 3 years ago
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    ya no help from there...google told me to come here haha

  25. FibonacciChick666
    • 3 years ago
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    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

  26. Best_Mathematician
    • 3 years ago
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    what lol...say it again confused

  27. FibonacciChick666
    • 3 years ago
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    h/o a minute

  28. Best_Mathematician
    • 3 years ago
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    ok ok

  29. FibonacciChick666
    • 3 years ago
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    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.

  30. Brotherman
    • 3 years ago
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    lol...no idea to this man.....

  31. Best_Mathematician
    • 3 years ago
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    haha...i am never gonna figure it out XD

  32. Brotherman
    • 3 years ago
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    I will copy it and show it to my lecturer and will get back to you

  33. Best_Mathematician
    • 3 years ago
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    sweet...dont forget to get the equation from comment box

  34. Brotherman
    • 3 years ago
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    yeah, I copied it

  35. Best_Mathematician
    • 3 years ago
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    thx tell me ASAP plz

  36. Best_Mathematician
    • 3 years ago
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    @amistre64 can u solve this

  37. amistre64
    • 3 years ago
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    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

  38. amistre64
    • 3 years ago
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    .... that i can determine

  39. Best_Mathematician
    • 3 years ago
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    haha...but there is an error idk wht

  40. ZeHanz
    • 3 years ago
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    @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...

  41. inkyvoyd
    • 3 years ago
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    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.

  42. FibonacciChick666
    • 3 years ago
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    @inkyvoyd ...that would make so much more sense

  43. Best_Mathematician
    • 3 years ago
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    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

  44. FibonacciChick666
    • 3 years ago
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    did you ask your instructor?

  45. inkyvoyd
    • 3 years ago
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    The error is that "an error exists" itself then, I believe. Please ask your instructor :)

  46. Best_Mathematician
    • 3 years ago
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    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

  47. Best_Mathematician
    • 3 years ago
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    just watch this question

  48. Edutopia
    • 3 years ago
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    There is no error in the OP. int(tanx)=int(sinx/cosx)=lnabs(secx)

  49. Edutopia
    • 3 years ago
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    maby your getting -ln(abs(cosx)) but using the properties of logarithms this equals ln(abs(cos^(-1)x)) = ln(abs(secx))

  50. Best_Mathematician
    • 3 years ago
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    wrong answer

  51. Best_Mathematician
    • 3 years ago
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    will reply the correct one asap

  52. amistre64
    • 3 years ago
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    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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