The indefinite integral of cosecant function with respect to $x$ is written mathematically in the following two forms in calculus.

$(1).\,\,\,$ $\displaystyle \int{\csc{x}}\,dx$

$(2).\,\,\,$ $\displaystyle \int{\operatorname{cosec}{x}}\,dx$

Now, let us learn how to derive the integral rule for cosecant function with understandable steps in integral calculus.

The integration of by parts method is not useful to find the integration of the cosecant function. However, its integration can be calculated by a special technique in the integral calculus.

$=\,\,\,$ $\displaystyle \int{(\csc{x} \times 1)}\,dx$

Write the factor $1$ as follows for our convenience. It means nothing is changed by multiplying the cosecant function with the subtraction of cot function from cosecant function, and then dividing their product of by the same expression.

$=\,\,\,$ $\displaystyle \int{\bigg(\csc{x} \times \dfrac{\csc{x}-\cot{x}}{\csc{x}-\cot{x}}\bigg)}\,dx$

It is time to find the product of the functions by multiplying the fractions.

$=\,\,\,$ $\displaystyle \int{\dfrac{\csc{x} \times (\csc{x}-\cot{x})}{\csc{x}-\cot{x}}}\,dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{\csc{x} \times \csc{x}-\csc{x} \times \cot{x}}{\csc{x}-\cot{x}}}\,dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{\csc^2{x}-\csc{x}\cot{x}}{\csc{x}-\cot{x}}}\,dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{-\csc{x}\cot{x}+\csc^2{x}}{\csc{x}-\cot{x}}}\,dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{(-\csc{x}\cot{x}+\csc^2{x}) \times dx}{\csc{x}-\cot{x}}}$

$=\,\,\,$ $\displaystyle \int{\dfrac{(\csc^2{x}-\csc{x}\cot{x}) \times dx}{\csc{x}-\cot{x}}}$

Take $v \,=\, \csc{x}-\cot{x}$

Now, differentiate the expressions on both sides of the equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(v)}$ $\,=\,$ $\dfrac{d}{dx}{(\csc{x}-\cot{x})}$

The derivative of the difference of the functions can be evaluated by the subtraction rule of the differentiation.

$\implies$ $\dfrac{dv}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(\csc{x})}$ $-$ $\dfrac{d}{dx}{(\cot{x})}$

Now, use the derivative rule of cosecant function and derivative rule of cot function to find the differentiations of them with respect to $x$ mathematically.

$\implies$ $\dfrac{dv}{dx}$ $\,=\,$ $-\csc{x}\cot{x}$ $-$ $(-\csc^2{x})$

$\implies$ $\dfrac{dv}{dx}$ $\,=\,$ $-\csc{x}\cot{x}$ $+$ $\csc^2{x}$

$\implies$ $\dfrac{dv}{dx}$ $\,=\,$ $\csc^2{x}$ $-$ $\csc{x}\cot{x}$

$\,\,\,\therefore\,\,\,\,\,\,$ $dv$ $\,=\,$ $(\csc^2{x}-\csc{x}\cot{x}) \times dx$

It is the right time to convert the whole integral function in terms of $x$ into the integral function in terms of $v$.

$\implies$ $\displaystyle \int{\dfrac{(\csc^2{x}-\csc{x}\cot{x}) \times dx}{\csc{x}-\cot{x}}}$ $\,=\,$ $\displaystyle \int{\dfrac{dv}{v}}$

Let us find the integration of the reciprocal of the variable $v$ with respect to $v$.

$=\,\,\,$ $\displaystyle \int{\dfrac{1 \times dv}{v}}$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{v}} \times dv$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{v}}\,dv$

Use the reciprocal integral rule to find the integration of the multiplicative inverse of the variable $v$ with respect to $v$.

$=\,\,\,$ $\log_{e}{|v|}+c$

Now, substitute the value of $v$ in terms of $x$ to finish the process of finding the integral of the cosecant function.

$=\,\,\,$ $\log_{e}{|\csc{x}-\cot{x}|}+c$

This natural logarithmic function can also be written in the following form as per the logarithms.

$=\,\,\,$ $\ln{|\csc{x}-\cot{x}|}+c$

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