# How do you evaluate the integral int arc cotx?

Jul 21, 2017

$\int \setminus a r c \cot x \setminus \mathrm{dx} = x \setminus a r c \cot x + \frac{1}{2} \setminus \ln \left({x}^{2} + 1\right) + C$

#### Explanation:

We seek:

$I = \int \setminus a r c \cot x \setminus \mathrm{dx}$

We can Integration By Parts (IBP):

Let $\left\{\begin{matrix}u & = a r c \cot x & \implies & \frac{\mathrm{du}}{\mathrm{dx}} & = - \frac{1}{{x}^{2} + 1} \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = 1 & \implies & v & = x\end{matrix}\right.$

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

gives us

$\int \setminus \left(a r c \cot x\right) \left(1\right) \setminus \mathrm{dx} = \left(a r c \cot x\right) \left(x\right) - \int \setminus \left(x\right) \left(- \frac{1}{{x}^{2} + 1}\right) \setminus \mathrm{dx}$

$\therefore I = x \setminus a r c \cot x + \int \setminus \frac{x}{{x}^{2} + 1} \setminus \mathrm{dx}$
$\text{ } = x \setminus a r c \cot x + \frac{1}{2} \setminus \int \setminus 2 \frac{x}{{x}^{2} + 1} \setminus \mathrm{dx}$
$\text{ } = x \setminus a r c \cot x + \frac{1}{2} \setminus \ln \left({x}^{2} + 1\right) + C$

Note

Normally when we integrate an integrand of the form $\frac{f ' \left(x\right)}{f \left(x\right)}$ we write the result as:

$\int \setminus \frac{f ' \left(x\right)}{f \left(x\right)} \setminus \mathrm{dx} = \ln | f \left(x\right) | \iff \int \setminus \frac{1}{u} \setminus \mathrm{du} = \ln | u |$

In the above problem the absolute signs are omitted as:

${x}^{2} + 1 > 0 \forall x \in \mathbb{R} \implies | {x}^{2} + 1 | = {x}^{2} + 1$

Thus we should write:

$I = x \setminus a r c \cot x + \frac{1}{2} \setminus \ln | {x}^{2} + 1 | + C$

But in this case this is equivalent to:

$I = x \setminus a r c \cot x + \frac{1}{2} \setminus \ln \left({x}^{2} + 1\right) + C$