1. Sep 10, 2009

### pillar

1.For example

a.http://img40.imageshack.us/img40/9514/problem2m.png [Broken]

b.http://img193.imageshack.us/img193/4250/problem2ans.png [Broken]

When do you use ln in integrals?

Last edited by a moderator: May 4, 2017
2. Sep 10, 2009

### HallsofIvy

One of the things you should have learned early is that
$$\int \frac{1}{x}dx= ln|x|+ C$$

$\int sin(x) dx$ 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.

3. Sep 10, 2009

### pillar

Sorry ln|cos(x)|+ C+ must be the integral of tan(x) then.

4. Sep 10, 2009

### HallsofIvy

Almost. If you let u= cos(x) then du= -sin(x)dx so
$$\int tan(x)dx= \int\frac{sin(x)}{cos(x)}dx= -\int \frac{du}{u}=-ln|u|+ C= -ln|cos(x)|+ C$$

5. Sep 10, 2009

### Staff: Mentor

BTW, the shorthand form is not In--it is ln, an abbreviation for logarithmus naturalis, Latin for natural logarithm.

6. Sep 10, 2009

### njama

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)

So $\int{tan(x)}=-ln|cos(x)|+C$.