When do you use In or log in Integrals?

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Homework Help Overview

The discussion revolves around the use of natural logarithms (ln) in integral calculus, specifically in relation to the integral of functions such as 1/x and tan(x). Participants are exploring when to apply ln in the context of integrals and clarifying misconceptions regarding integral results.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the integral of 1/x and its result involving ln, while questioning the validity of other suggested integral forms, such as ln|cos(x)|. There is an exploration of the relationship between derivatives and integrals, particularly in the context of tan(x).

Discussion Status

The discussion is active with participants providing clarifications and corrections regarding the use of ln in integrals. Some participants are questioning assumptions and definitions, while others are attempting to clarify the correct forms of integrals and their derivatives.

Contextual Notes

There appears to be some confusion regarding the notation and terminology used for natural logarithms, as well as the specific integrals being discussed. Participants are navigating through these misunderstandings while adhering to the constraints of the homework context.

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1.For example

a.http://img40.imageshack.us/img40/9514/problem2m.png

b.http://img193.imageshack.us/img193/4250/problem2ans.png
When do you use ln in integrals?
 
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One of the things you should have learned early is that
[tex]\int \frac{1}{x}dx= ln|x|+ C[/tex]

[itex]\int sin(x) dx[/itex] is, as you first say, - cos(x)+ C. It is certainly NOT equal to "ln|cos(x)|+ C. I don't know why you would even suggest that.
 
HallsofIvy said:
One of the things you should have learned early is that
[tex]\int \frac{1}{x}dx= ln|x|+ C[/tex]

[itex]\int sin(x) dx[/itex] is, as you first say, - cos(x)+ C. It is certainly NOT equal to "ln|cos(x)|+ C. I don't know why you would even suggest that.

Sorry ln|cos(x)|+ C+ must be the integral of tan(x) then.
 
pillar said:
Sorry ln|cos(x)|+ C+ must be the integral of tan(x) then.
Almost. If you let u= cos(x) then du= -sin(x)dx so
[tex]\int tan(x)dx= \int\frac{sin(x)}{cos(x)}dx= -\int \frac{du}{u}=-ln|u|+ C= -ln|cos(x)|+ C[/tex]
 
BTW, the shorthand form is not In--it is ln, an abbreviation for logarithmus naturalis, Latin for natural logarithm.
 
Just find the derivative of ln|cos(x)| and you will be sure, what the correct integral is.

[-ln|cos(x)| ]'=(-ln(u))'u'=(-1/cos(x))*(-sin(x))=sin(x)/cos(x)=tan(x) :smile:

So [itex]\int{tan(x)}=-ln|cos(x)|+C[/itex].
 

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