Simple Trigonometric Integral

1. Jul 2, 2013

eddybob123

Hi, I am just wondering what the antiderivative of this integral was. It looks easy to me, but I no matter what I did just could not evaluate it. Can someone help me?:
$$\int \frac{\tan(x)}{x}\;dx$$

2. Jul 2, 2013

MarneMath

Have you tried integration by parts? It can be a bit tedious but it isn't to bad.

3. Jul 4, 2013

TheFool

Actually, that integral isn't so simple. It cannot be stated as a finite combination of elementary functions.

Last edited: Jul 4, 2013
4. Jul 4, 2013

Goa'uld

In fact, that is true for all trig functions over x.

5. Jul 5, 2013

eddybob123

How about using the Fourier series for tan?

6. Jul 6, 2013

Goa'uld

It depends on the application. If this is part of a larger problem it may be more convenient to define a new function in terms of the integral rather than write an infinite series repeatedly.
$$f'(x)=\frac{\tan{x}}{x}$$
$$f(x)=\int_{0}^{x}\frac{\tan{t}}{t}dt$$
One problem that may occur with infinite series is the radius of convergence. While tan(x)/x may have a value at certain values of x, the infinite series may not converge.

Last edited: Jul 6, 2013
7. Jul 6, 2013

babysnatcher

Use integration by parts twice.

8. Jul 6, 2013

9. Jul 6, 2013

MarneMath

Bleh, let dv = 1/x u = tanx for the first integration by parts.

Then with the second integral you get let dv = 1/x and dv = sec^2(x)dx. You'll notice that after you do integration by parts twice, that you'll have the same integral on both sides, so combine like terms and come out with ln(x)tan(x)

10. Jul 6, 2013

Goa'uld

You are mistaken.
$$\frac{d}{dx}(\log{x}\tan{x})=\log{x}\sec^{2}{x}+\frac{\tan{x}}{x}$$

Here is where you went wrong.
$$\int\frac{\tan{x}}{x}dx=\log{|x|}\tan{x}-\int\log{|x|}\sec^{2}{x}dx$$

Last edited: Jul 6, 2013
11. Jul 9, 2013

DeeAytch

What would this look like?

12. Jul 9, 2013