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## Best_Mathematician 2 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

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

2. 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?

3. Best_Mathematician

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

4. Best_Mathematician

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

5. 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.

6. Best_Mathematician

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

7. P0sitr0n

no brackets should be integral (sinx/cosx)dx

8. Best_Mathematician

what

9. P0sitr0n

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

10. Best_Mathematician

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

11. P0sitr0n

lol just guessing maybe its this

12. Best_Mathematician

dude i m frikin out lol

13. 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

14. FibonacciChick666

correction $-ln|cosx|+C$

15. Best_Mathematician

ya so whats the error

16. 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 $-ln|cos x|+C$

17. Best_Mathematician

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

18. 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

19. Best_Mathematician

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

20. 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

21. Best_Mathematician

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

22. Best_Mathematician

haha

23. FibonacciChick666

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

24. Best_Mathematician

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

25. 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

26. Best_Mathematician

what lol...say it again confused

27. FibonacciChick666

h/o a minute

28. Best_Mathematician

ok ok

29. 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.

30. Brotherman

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

31. Best_Mathematician

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

32. Brotherman

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

33. Best_Mathematician

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

34. Brotherman

yeah, I copied it

35. Best_Mathematician

thx tell me ASAP plz

36. Best_Mathematician

@amistre64 can u solve this

37. 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

38. amistre64

.... that i can determine

39. Best_Mathematician

haha...but there is an error idk wht

40. 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...

41. 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.

42. FibonacciChick666

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

43. Best_Mathematician

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

did you ask your instructor?

45. inkyvoyd

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

46. Best_Mathematician

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

just watch this question

48. Edutopia

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

49. Edutopia

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

wrong answer

51. Best_Mathematician

will reply the correct one asap

52. 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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