Integral of $$$c \cot{\left(x \right)}$$$ with respect to $$$x$$$
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Find $$$\int c \cot{\left(x \right)}\, dx$$$.
Solution
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=c$$$ and $$$f{\left(x \right)} = \cot{\left(x \right)}$$$:
$${\color{red}{\int{c \cot{\left(x \right)} d x}}} = {\color{red}{c \int{\cot{\left(x \right)} d x}}}$$
Rewrite the cotangent as $$$\cot\left(x\right)=\frac{\cos\left(x\right)}{\sin\left(x\right)}$$$:
$$c {\color{red}{\int{\cot{\left(x \right)} d x}}} = c {\color{red}{\int{\frac{\cos{\left(x \right)}}{\sin{\left(x \right)}} d x}}}$$
Let $$$u=\sin{\left(x \right)}$$$.
Then $$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (steps can be seen »), and we have that $$$\cos{\left(x \right)} dx = du$$$.
Therefore,
$$c {\color{red}{\int{\frac{\cos{\left(x \right)}}{\sin{\left(x \right)}} d x}}} = c {\color{red}{\int{\frac{1}{u} d u}}}$$
The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$c {\color{red}{\int{\frac{1}{u} d u}}} = c {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
Recall that $$$u=\sin{\left(x \right)}$$$:
$$c \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = c \ln{\left(\left|{{\color{red}{\sin{\left(x \right)}}}}\right| \right)}$$
Therefore,
$$\int{c \cot{\left(x \right)} d x} = c \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)}$$
Add the constant of integration:
$$\int{c \cot{\left(x \right)} d x} = c \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)}+C$$
Answer
$$$\int c \cot{\left(x \right)}\, dx = c \ln\left(\left|{\sin{\left(x \right)}}\right|\right) + C$$$A