Integral of $$$\cot{\left(c \right)}$$$
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Find $$$\int \cot{\left(c \right)}\, dc$$$.
Solution
Rewrite the cotangent as $$$\cot\left(c\right)=\frac{\cos\left(c\right)}{\sin\left(c\right)}$$$:
$${\color{red}{\int{\cot{\left(c \right)} d c}}} = {\color{red}{\int{\frac{\cos{\left(c \right)}}{\sin{\left(c \right)}} d c}}}$$
Let $$$u=\sin{\left(c \right)}$$$.
Then $$$du=\left(\sin{\left(c \right)}\right)^{\prime }dc = \cos{\left(c \right)} dc$$$ (steps can be seen »), and we have that $$$\cos{\left(c \right)} dc = du$$$.
Therefore,
$${\color{red}{\int{\frac{\cos{\left(c \right)}}{\sin{\left(c \right)}} d c}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$
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 $$$u=\sin{\left(c \right)}$$$:
$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\sin{\left(c \right)}}}}\right| \right)}$$
Therefore,
$$\int{\cot{\left(c \right)} d c} = \ln{\left(\left|{\sin{\left(c \right)}}\right| \right)}$$
Add the constant of integration:
$$\int{\cot{\left(c \right)} d c} = \ln{\left(\left|{\sin{\left(c \right)}}\right| \right)}+C$$
Answer
$$$\int \cot{\left(c \right)}\, dc = \ln\left(\left|{\sin{\left(c \right)}}\right|\right) + C$$$A