Integration of cot x

Integration of cot x can be found using various integration technique like integration by substitution along with trigonometric identities. The formula for integration of cot x is

\[
\int \cot(x) \, dx = \ln |sin(x)| + C
\]

Proof of the Integration of cot x

Integration of cot x can be solved using integration by substitution as given below

$\int \frac {f^{‘} (x)}{f(x)} \; dx = ln | f(x)| + C$
Proof
let us substitute
f(x) =t
then
f'(x) dx=dt
Therefore
$\int \frac {f^{‘} (x)}{f(x)}\; dx = \int \frac {1}{t} \; dt =ln |t| + C= ln | f(x)| + C $

Now

\[
\int \cot(x) \, dx = \int \frac {\cos(x)}{\sin(x)} \, dx
\]

Now
$\sin(x) =t$
then
$\cos (x) dx = dt$
Therefore,

\[
\int \cot(x) \, dx = \int \frac {1}{t} \, dt = \ln |sin(x)| + C
\]

Definite Integral of cot x

To evaluate the definite integral of $\cot(x)$ over a specific interval, say from (a) to (b), you follow the same process as for the indefinite integral, but then apply the limits of integration. The indefinite integral of $\cot(x)$ is $\ln|\sin(x)| + C$.

So, the definite integral

\[
\int_{a}^{b} \cot(x) \, dx = \left[ \ln|\sin(x)| \right]_{a}^{b} = \ln|\sin(b)| – \ln|\sin(a)|.
\]

However, it’s important to be cautious about the interval of integration because $\cot(x)$ and $\ln|\sin(x)|$ have singularities (points where they are not defined). Specifically, $\cot(x)$ and $\ln|\sin(x)|$ are undefined where $\sin(x) = 0$, which occurs at integer multiples of $\pi$. Therefore, the interval ([a, b]) should not include points where $\sin(x) = 0$

Example

\[
\int_{0}^{\pi/2} \cot(x) \, dx = \left[ \ln|\sin(x)| \right]_{0}^{\pi/2} = \ln|\sin(\pi/2)| – \ln|\sin(0)| = \ln(1) – \ln(0).
\]

However, this integral is problematic at (x = 0) because $\ln|\sin(0)|$ is undefined (as $\ln(0)$ is not defined). In such cases, the integral does not have a standard value

Solved Examples

Question 1

Find

\[
\int_{\pi/4}^{\pi/2} \cot(x) \, dx
\]

Solution

\[
\int_{\pi/4}^{\pi/2} \cot(x) \, dx = \left[ \ln|\sin(x)| \right]_{\pi/4}^{\pi/2}.
\]

\[
\begin{align} \int_{\pi/4}^{\pi/2} \cot(x) \, dx &= \ln|\sin(\pi/2)| – \ln|\sin(\pi/4)| \\ &= \ln(1) – \ln\left(\frac{1}{\sqrt{2}}\right) \\ &= 0 – \ln\left(\frac{1}{\sqrt{2}}\right) \\ &= -\ln\left(\frac{1}{\sqrt{2}}\right). \end{align}
\]

Since $\frac{1}{\sqrt{2}}$ is the sine of $\pi/4$, this result is valid. Also, note that $\ln\left(\frac{1}{\sqrt{2}}\right)$ can be simplified further:

\[
-\ln\left(\frac{1}{\sqrt{2}}\right) = \ln\left(\sqrt{2}\right),
\]

Therefore, the definite integral is $\ln\left(\sqrt{2}\right)$.