Integral of Inverse Cosecant

The integral of inverse cosecant {\csc ^{ - 1}}x it is an important integral function, but it has no direct method to find. We shall find the integration of inverse cosecant by using the integration by parts method.

The integral of inverse cosecant is of the form

I = \int {{{\csc }^{ - 1}}xdx}

To solve this integration it must have at least two functions, however it has only one function: {\csc ^{ - 1}}x. So consider second function as 1. Now the integration becomes

I = \int {{{\csc }^{ - 1}}x \cdot 1dx} \,\,\,\,{\text{ - - - }}\left( {\text{i}} \right)

The first function is {\csc ^{ - 1}}x and the second function is 1

Using the formula for integration by parts, we have

\int {\left[ {f\left( x \right)g\left( x \right)} \right]dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left[ {\frac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } }

Using the formula above, equation (i) becomes

\begin{gathered} I = {\csc ^{ - 1}}x\int {1dx - \int {\left[ {\frac{d}{{dx}}{{\csc }^{ - 1}}x\int {1dx} } \right]dx} } \\ \Rightarrow I = x{\csc ^{ - 1}}x - \int {\left[ { - \frac{1}{{x\sqrt {{x^2} - 1} }}x} \right]dx} \\ \Rightarrow I = x{\csc ^{ - 1}}x + \int {\frac{1}{{\sqrt {{x^2} - 1} }}dx} \\ \end{gathered}

It can be written in the form

I = x{\csc ^{ - 1}}x + \int {\frac{1}{{\sqrt {{x^2} - {{\left( 1 \right)}^2}} }}dx}

Using formula

\int {\frac{1}{{\sqrt {{x^2} - {a^2}} }}dx = {{\cosh }^{ - 1}}\left( {\frac{x}{a}} \right)} + c

we have

\begin{gathered} I = x{\csc ^{ - 1}}x + {\cosh ^{ - 1}}\left( {\frac{x}{1}} \right) + c \\ \Rightarrow \int {{{\csc }^{ - 1}}xdx} = x{\csc ^{ - 1}}x + {\cosh ^{ - 1}}x + c \\ \end{gathered}

Now we can also use this integration of inverse cosecant as a formula.