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

1 Answer
Jul 21, 2017

# 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 #