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

Question

Cameron Alexander · Accepted Answer

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

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 #

Christopher Agresta · Answer

undefined To evaluate the integral of $\int \arccot(x) \, dx$, you can use integration by parts. Let $u = \arccot(x)$ and $dv = dx$. Then, $du = -\frac{1}{1+x^2} \, dx$ and $v = x$. Applying the integration by parts formula:

$$
\begin{align*}
\int \arccot(x) \, dx &= uv - \int v \, du \
&= x \arccot(x) - \int x \left(-\frac{1}{1+x^2}\right) \, dx \
&= x \arccot(x) + \int \frac{x}{1+x^2} \, dx.
\end{align*}
$$

The integral $\int \frac{x}{1+x^2} \, dx$ can be evaluated by a simple substitution. Let $u = 1+x^2$, then $du = 2x \, dx$. Thus:

$$
\begin{align*}
\int \frac{x}{1+x^2} \, dx &= \frac{1}{2} \int \frac{du}{u} \
&= \frac{1}{2} \ln|u| + C \
&= \frac{1}{2} \ln|1+x^2| + C.
\end{align*}
$$

Therefore, the integral $\int \arccot(x) \, dx$ simplifies to:

$$
x \arccot(x) + \frac{1}{2} \ln|1+x^2| + C,
$$

where $C$ is the constant of integration.