Question

What is $\int{arc\cot x\,dx}$ ?

Answer 1

Hint: Here we have to integrate the given value. Firstly as we can see that we have been given a trigonometric function we will take another value as $1$ . Then by using ILATE Rule we will take the first and second function from the integral. Finally by using by-parts method given as $\int{uv\,dx}=u\int{v\,dx}-\int{\left( \dfrac{du}{dx}\int{v\,dx} \right)dx}$ we will find the value of the integral and get our desired answer.

Complete step-by-step solution:
We have to solve the below integral:
$\int{arc\cot x\,dx}$
As we know $arc\cot x={{\cot }^{-1}}x$ we will replace the value above.
$\int{arc\cot x\,dx}=\int{{{\cot }^{-1}}x\,dx}$…$\left( 1 \right)$
Now we can write the above value as follows:
$\int{{{\cot }^{-1}}x\,dx}=\int{1\times {{\cot }^{-1}}x\,dx}$…$\left( 2 \right)$
We will be using By-parts method to solve the above integral given as follows:
$\int{uv\,dx}=u\int{v\,dx}-\int{\left( \dfrac{du}{dx}\int{v\,dx} \right)dx}$….$\left( 3 \right)$
Now for using the above method we want our two functions.
Using ILATE (Inverse, Algebraic, Logarithm, Trigonometric, Exponent) Rule
So by the above Rule our first function will be the inverse function and the second function will be the Algebraic number.
So from equation (2) we have the value of two functions as:
$u={{\cot }^{-1}}x$ And $v=1$
Substitute the above value in equation (3) as follows:
$\Rightarrow \int{{{\cot }^{-1}}x\times 1\,dx}={{\cot }^{-1}}x\int{1\,dx}-\int{\left( \dfrac{d\left( {{\cot }^{-1}}x \right)}{dx}\int{1\,dx} \right)dx}$
Now we know integration of $\int{1\,dx}=x$ and $\dfrac{d\left( {{\cot }^{-1}}x \right)}{dx}=-\dfrac{1}{{{x}^{2}}+1}$ substitute the value above,
$\Rightarrow \int{{{\cot }^{-1}}x\times 1\,dx}={{\cot }^{-1}}x\times x-\int{\left( -\dfrac{1}{{{x}^{2}}+1}\times x \right)dx}$
$\Rightarrow \int{{{\cot }^{-1}}x\times 1\,dx}=x{{\cot }^{-1}}x+\int{\dfrac{x}{{{x}^{2}}+1}dx}$…$\left( 4 \right)$
Now we will use substitution method on the integral on the right as follows:
Let $u={{x}^{2}}+1$…$\left( 5 \right)$
Then on differentiating above value we get,
$\Rightarrow \dfrac{du}{dx}=2x$
$\Rightarrow \dfrac{du}{2}=x\,dx$…$\left( 6 \right)$
Put the value from equation (5) and (6) in equation (4) we get,
$\Rightarrow \int{{{\cot }^{-1}}x\times 1\,dx}=x{{\cot }^{-1}}x+\int{\dfrac{1}{u}\times \dfrac{du}{2}}$
Now as we know $\int{\dfrac{1}{x}dx=\log x+C}$ where $C$ is any constant using it above we get,
$\Rightarrow \int{{{\cot }^{-1}}x\times 1\,dx}=x{{\cot }^{-1}}x+\dfrac{1}{2}\log u+C$
Replace the value from equation (5) and (2) above we get,
$\Rightarrow \int{{{\cot }^{-1}}x\,dx}=x{{\cot }^{-1}}x+\dfrac{1}{2}\log \left( {{x}^{2}}+1 \right)+C$
Put the above value in equation (1) we get,
$\Rightarrow \int{arc\cot x\,dx}=x{{\cot }^{-1}}x+\dfrac{1}{2}\log \left( {{x}^{2}}+1 \right)+C$
Also we know $arc\cot x={{\cot }^{-1}}x$ so,
$\Rightarrow \int{arc\cot x\,dx}=x\,arc\cot x+\dfrac{1}{2}\log \left( {{x}^{2}}+1 \right)+C$
Hence $\int{arc\cot x\,dx}=x\,arc\cot x+\dfrac{1}{2}\log \left( {{x}^{2}}+1 \right)+C$ where $C$ is any constant.

Note: Integration is one of the two important topics of calculus. It is used to find the area, volume and other values that occur when a collection of data is taken. In this question s we have been given an inverse function inside the integral sign it is not possible to solve it directly that is why we have taken a constant $1$ and have used integration by-parts. ILATE Rule is very important for using the by-parts method for taking which function will be treated as first and which will be treated as second.