Assume, $x$ as a variable and represents angle of a right triangle. In trigonometry, the cosecant squared of angle $x$ is written as $\csc^2{x}$ or $\operatorname{cosec}^2{x}$ in mathematical form. The indefinite integration of cosecant squared function with respect to $x$ is written as the following mathematical form in differential calculus

$\displaystyle \int{\csc^2{x} \,}dx \,\,\,$ (or) $\,\,\, \displaystyle \int{\operatorname{cosec}^2{x} \,}dx$

Now, let us start deriving the integration formula for the cosecant squared function in integral calculus.

Express the derivative of cot function formula with respect to $x$ in mathematical form.

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

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

As per differential calculus, the derivative of a constant is always zero. So, there is no problem in adding an arbitrary constant to cot function.

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

According to integral calculus, the collection of all primitives of $\csc^2{x}$ function is called the integration of $\csc^2{x}$ function. It can be written in mathematical form in two ways.

$\displaystyle \int{\csc^2{x} \,}dx \,\,\,$ (or) $\,\,\, \displaystyle \int{\operatorname{cosec}^2{x} \,}dx$

The primitive or an antiderivative of $\csc^2{x}$ function is $-\cot{x}$ and the constant of integration ($c$) in this case.

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

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

Therefore, it has proved that the indefinite integration of cosecant squared of an angle function is equal to the sum of the negative cot function and a constant of integration.

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