Question

How do you evaluate the integral `int arc cotx`?

Answer 1

`int \ arc cot x \ dx =x \ arc cotx + 1/2 \ ln (x^2+1) + C`

Explanation:

We seek:

`I = int \ arc cot x \ dx`

We can Integration By Parts (IBP):

Let `{ (u,=arc cotx, => , (du)/dx,=-1/(x^2+1)), ((dv)/dx,=1, =>, v,=x ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

gives us

`int \ (arc cotx)(1) \ dx = (arc cotx)(x) - int \ (x)(-1/(x^2+1)) \ dx`

`:. I = x \ arc cotx + int \ x/(x^2+1) \ dx`
`" " = x \ arc cotx + 1/2 \ int \ 2x/(x^2+1) \ dx`
`" " = x \ arc cotx + 1/2 \ ln (x^2+1) + C`

Note

Normally when we integrate an integrand of the form `(f'(x))/(f(x))` we write the result as:

`int \ (f'(x))/(f(x)) \ dx = ln |f(x)| iff int \ 1/u \ du = ln |u|`

In the above problem the absolute signs are omitted as:

`x^2+1 gt 0 AA x in RR => |x^2+1| = x^2+1`

Thus we should write:

`I = x \ arc cotx + 1/2 \ ln |x^2+1| + C`

But in this case this is equivalent to:

`I = x \ arc cotx + 1/2 \ ln (x^2+1) + C`