Question

How do you find the antiderivative of the function $y=\ arc cot x$ ?

Answer 1

Hint: We have been given an inverse of a trigonometric function whose antiderivative is to be found. From our prior knowledge of basic calculus, we know that the antiderivative is another name for the integral of the function. Thus, we shall integrate the given trigonometric function using the ILATE rule. Further, we shall use substitution to simplify the integration by applying the integration by parts.

Complete step by step solution:
Given that $y=\ arc cot x$.
The symbolic mathematical representation of $\ arc cot x$ is $co{{t}^{-1}}x$ as it represents the inverse of cotangent function.
$\Rightarrow y=co{{t}^{-1}}x$
This can be also written as $y=\left( 1 \right)co{{t}^{-1}}x$ because the product of any function with 1 is the function itself.
$\Rightarrow y=\left( 1 \right)co{{t}^{-1}}x$
Here, we shall apply integration by parts. Here, our first function is 1 and our second function is $co{{t}^{-1}}x$.
\[\Rightarrow \int{\left( 1 \right)co{{t}^{-1}}x.dx=co{{t}^{-1}}x.\int{1.dx}}-\int{\left( \int{1.dx} \right).\left( \dfrac{d}{dx}co{{t}^{-1}}x \right)}.dx\]
We know by the formulae of differentiation that $\dfrac{d}{dx}{{\cot }^{-1}}x=-\dfrac{1}{1+{{x}^{2}}}$ and by the formulae of integration that $\int{1.dx=x+C}$.
 \[\Rightarrow \int{\left( 1 \right)co{{t}^{-1}}x.dx=co{{t}^{-1}}x\left( x+C \right)}-\int{\left( x \right).\left( -\dfrac{1}{1+{{x}^{2}}} \right)}.dx\]
\[\Rightarrow \int{\left( 1 \right)co{{t}^{-1}}x.dx=xco{{t}^{-1}}x}-\int{\left( -\dfrac{x}{1+{{x}^{2}}} \right)}.dx\]
We shall make use of substitution here to simplify our integration.
Let $t=1+{{x}^{2}}$
Differentiating both sides, we get
$\Rightarrow dt=2x.dx$
Applying these substitutions, we get
\[\Rightarrow \int{\left( 1 \right)co{{t}^{-1}}x.dx=xco{{t}^{-1}}x}+\int{\dfrac{1}{t}}.\dfrac{dt}{2}\]
\[\Rightarrow \int{\left( 1 \right)co{{t}^{-1}}x.dx=xco{{t}^{-1}}x}+\dfrac{1}{2}\int{\dfrac{1}{t}}.dt\]
\[\Rightarrow \int{\left( 1 \right)co{{t}^{-1}}x.dx=xco{{t}^{-1}}x}+\dfrac{1}{2}\ln t+C\]
Now, we shall substitute the value of variable-t.
\[\Rightarrow \int{\left( 1 \right)co{{t}^{-1}}x.dx=xco{{t}^{-1}}x}+\dfrac{1}{2}\ln \left( 1+{{x}^{2}} \right)+C\]
\[\Rightarrow \int{co{{t}^{-1}}x.dx=xco{{t}^{-1}}x}+\dfrac{1}{2}\ln \left( 1+{{x}^{2}} \right)+C\]
Therefore, the antiderivative of the function $y=\ arc cot x$ is \[xco{{t}^{-1}}x+\dfrac{1}{2}\ln \left( 1+{{x}^{2}} \right)+C\].

Note: Since we do not have a direct formula of integration for the inverse of cotangent function, therefore, we integrated this function by applying integration by parts. Here, the function was manipulated by writing it as the product of 1 and the function itself. In this manner, it could be made ready to undergo integration by parts.