When do you use In or log in Integrals?

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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]
 
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].