# How do you find the antiderivative of the function y=arc cotx?

Sep 3, 2015

The antiderivative or integral of this is:

int "arccot"xdx = color(blue)(x"arccot"x + 1/2ln(1+x^2) + C)

You tend to have to do inverse trig integrals using Integration by Parts (try it on $\int \arctan x \mathrm{dx}$ sometime).

Let:
$u = \text{arccot} x$
$\mathrm{du} = - \frac{1}{1 + {x}^{2}} \mathrm{dx}$
$\mathrm{dv} = 1 \mathrm{dx}$
$v = x$

$u v - \int v \mathrm{du}$

$= x \text{arccot} x - \int \frac{- x}{1 + {x}^{2}} \mathrm{dx}$

$= x \text{arccot} x + \int \frac{x}{1 + {x}^{2}} \mathrm{dx}$

Now, you can determine the integral here using u-substitution. Let:
$u = 1 + {x}^{2}$
$\mathrm{du} = 2 x \mathrm{dx}$

$\implies \frac{1}{2} \int \frac{2 x}{1 + {x}^{2}} \mathrm{dx}$

$= \frac{1}{2} \int \frac{1}{u} \mathrm{du}$

$= \frac{1}{2} \ln | u | = \frac{1}{2} \ln | 1 + {x}^{2} | = \frac{1}{2} \ln \left(1 + {x}^{2}\right)$
(since $1 + {x}^{2} > 0$, always, for $x \in \mathbb{R}$)

int "arccot"xdx = color(blue)(x"arccot"x + 1/2ln(1+x^2) + C)
You can remember the derivative of $\text{arccot} x$ by simply remembering the derivative of $\arctan x$ and multiplying that by $- 1$. Simple pattern: take all the non-$c o \text{-}$ inverse trig functions and differentiate them. Then multiply by $- 1$ to get their $c o \text{-}$ counterparts, i.e. $\text{arccsc}$, $\text{arccot}$, and $\text{arccos}$ vs. $\text{arcsec}$, $\text{arctan}$, and $\text{arcsin}$, respectively.