Question

How do you evaluate `int cot(x)` from 0 to 2pi?

Answer 1

The integral diverges.

Explanation:

Consider the function: `cot(x)` has vertical asymptotes at `x=0,pi,2pi...` Thus the integral will diverge.

Answer 2

This improper integral does not converge.

Explanation:

`cot x` is not defined for `x = 0`, `pi` and `2pi`.

To attempt to evaluate the integral, we need to find the two improper integrals

`int_0^pi cot x dx + int_pi^(2pi) cotx dx`

`int_0^pi cot x dx` is undefined at both limits of integration, so we need to break it somewhere. I'll use `pi/2`

`int_0^pi cot x dx = int_0^(pi/2) cot x dx+int_(pi/2)^pi cot x dx`

`int_0^(pi/2) cot x dx = lim_(ararr0^+) int_a^(pi/2) cotx dx`

`= lim_(ararr0^+) ln(sinx) |_a^(pi/2)`

`= ln(sin(pi/2)) - lim_(ararr0^+) (ln(sina))`

`= ln(1) - lim_(ararr0^+) (ln(sina))`

`= - lim_(ararr0^+) (ln(sina))`

But this limit does not exist, so the integral diverges.

Because one of the integrals involved in `int_0^(2pi) cot x dx` diverges, the integral diverges.