Integral of $$$\cot{\left(t \right)}$$$

Find $$$\int \cot{\left(t \right)}\, dt$$$.

Rewrite the cotangent as $$$\cot\left(t\right)=\frac{\cos\left(t\right)}{\sin\left(t\right)}$$$:

$${\color{red}{\int{\cot{\left(t \right)} d t}}} = {\color{red}{\int{\frac{\cos{\left(t \right)}}{\sin{\left(t \right)}} d t}}}$$

Let $$$u=\sin{\left(t \right)}$$$.

Then $$$du=\left(\sin{\left(t \right)}\right)^{\prime }dt = \cos{\left(t \right)} dt$$$ (steps can be seen »), and we have that $$$\cos{\left(t \right)} dt = du$$$.

The integral can be rewritten as

$${\color{red}{\int{\frac{\cos{\left(t \right)}}{\sin{\left(t \right)}} d t}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$${\color{red}{\int{\frac{1}{u} d u}}} = {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recall that $$$u=\sin{\left(t \right)}$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\sin{\left(t \right)}}}}\right| \right)}$$

Therefore,

$$\int{\cot{\left(t \right)} d t} = \ln{\left(\left|{\sin{\left(t \right)}}\right| \right)}$$

Add the constant of integration:

$$\int{\cot{\left(t \right)} d t} = \ln{\left(\left|{\sin{\left(t \right)}}\right| \right)}+C$$

$$$\int \cot{\left(t \right)}\, dt = \ln\left(\left|{\sin{\left(t \right)}}\right|\right) + C$$$A