Math Doubts

Proof of Integral of cscx.cotx formula

$x$ is a variable and also denotes an angle of a right triangle. The cosecant and cotangent functions are written as $\csc{x}$ or $\operatorname{cosec}{x}$ and $\cot{x}$ respectively. The indefinite integration of product of cosecant and cot functions with respect to $x$ is written in the following mathematical form in integral calculus.

$\displaystyle \int{\csc{x}\cot{x} \,}dx$

Now, let’s derive the proof for the indefinite integration of product of $\csc{x}$ and $\cot{x}$ functions with respect to $x$ in integral calculus.

Derivative of Cosecant function

It is proved that the derivative of cosecant function is equal to the negative of product of cosecant and cotangent functions.

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

Inclusion of an Arbitrary constant

In fact, the derivative of a constant is zero. So, the derivative of cosecant function is same even though a constant is added to the cosecant function in differentiation. Add an arbitrary constant to cosecant function and now, differentiate the trigonometric function with respect to $x$.

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

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

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

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

$\implies$ $\dfrac{d}{dx}{(-\csc{x}+c_{1})} \,=\, \csc{x}\cot{x}$

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

Integral of cscx.cotx function

The collection of all primitives of product of $\csc{x}$ and $\cot{x}$ function is called the integration of product of cosecant and cot functions. As per the integral calculus, It can be written in mathematical form as follows.

$\displaystyle \int{\csc{x}\cot{x} \,}dx$

The primitive or an antiderivative of $\csc{x}\cot{x}$ function is equal to the sum of the $-\csc{x}$ function and the constant of integration ($c$).

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

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

Therefore, it is proved that the indefinite integration of product of cosecant and cot functions with respect to a variable is equal to the sum of the negative cosecant function and integral constant.



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