# Integral of Cotangent Inverse

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

The integral of cotangent inverse is of the form

\[I = \int {{{\cot }^{ – 1}}xdx} \]

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

\[I = \int {{{\cot }^{ – 1}}x \cdot 1dx} \,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)\]

The first function is $${\cot ^{ – 1}}x$$ 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

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

Multiplying and dividing by 2, we have

\[I = x{\cot ^{ – 1}}x + \frac{1}{2}\int {\frac{{2x}}{{1 + {x^2}}}dx} \]

Using formula \[\int {\frac{{f’\left( x \right)}}{{f\left( x \right)}}dx = \ln f\left( x \right)} + c\] we have

\[\begin{gathered} I = x{\cot ^{ – 1}}x + \frac{1}{2}\ln \left( {1 + {x^2}} \right) + c \\ \Rightarrow \int {{{\cot }^{ – 1}}xdx} = x{\cot ^{ – 1}}x + \frac{1}{2}\ln \left( {1 + {x^2}} \right) + c \\ \end{gathered} \]

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