Question

Evaluate the value of the limit $\int_{-2\pi }^{5\pi }{{{\cot }^{-1}}\left( \tan x \right)dx}$.

Answer 1

Hint: We here need to find the value of the definite integral $\int_{-2\pi }^{5\pi }{{{\cot }^{-1}}\left( \tan x \right)dx}$. We will here convert ${{\cot }^{-1}}\left( \tan x \right)$ into ${{\tan }^{-1}}\left( \tan x \right)$ by using the property ${{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}$ and then by using the property ${{\tan }^{-1}}\left( \tan x \right)=x$, we will bring the integral in the form of constants and x. then we will use the property $\int_{a}^{b}{{{x}^{n}}dx}=\left[ \dfrac{{{x}^{n+1}}}{n+1} \right]_{a}^{b}=\left[ \dfrac{{{b}^{n+1}}}{n+1}-\dfrac{{{a}^{n+1}}}{n+1} \right]$ and after solving via this, we will get our required answer.

Complete step by step answer:
Now, we have to find the value of the definite integral given as $\int_{-2\pi }^{5\pi }{{{\cot }^{-1}}\left( \tan x \right)dx}$. Let this be ‘I’.
Thus, we can say that:
$I=\int_{-2\pi }^{5\pi }{{{\cot }^{-1}}\left( \tan x \right)dx}$
Now, we will solve this using properties of trigonometry and that of definite integration.
Now, inside the integral sign, we have been given the function as ${{\cot }^{-1}}\left( \tan x \right)$ .
We will try to make it in the form of ${{\tan }^{-1}}\left( \tan x \right)$ because of the property given by:
${{\tan }^{-1}}\left( \tan x \right)=x\text{ }\forall x\in R$
Now, we know the property given as:
${{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}$
Now, if we replace ‘x’ in the argument of these functions with ${{\tan }^{-1}}\left( \tan x \right)$ , we will get the following result:
${{\cot }^{-1}}\left( \tan x \right)+{{\tan }^{-1}}\left( \tan x \right)=\dfrac{\pi }{2}$
$\Rightarrow {{\cot }^{-1}}\left( \tan x \right)=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( \tan x \right)$ …..(i)
Now, as mentioned above, we know that ${{\tan }^{-1}}\left( \tan x \right)=x$. Thus, we can write ${{\cot }^{-1}}\left( \tan x \right)$ as:
$\begin{align}
  & {{\cot }^{-1}}\left( \tan x \right)=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( \tan x \right) \\
 & \Rightarrow {{\cot }^{-1}}\left( \tan x \right)=\dfrac{\pi }{2}-x \\
\end{align}$
Thus, we get that \[{{\cot }^{-1}}\left( \tan x \right)=\dfrac{\pi }{2}-x\] …..(ii)
Now, we can substitute the value of ${{\cot }^{-1}}\left( \tan x \right)$ from equation (ii).
Substituting the value of ${{\cot }^{-1}}\left( \tan x \right)$ from equation (ii) in I we get:
$\begin{align}
  & I=\int_{-2\pi }^{5\pi }{{{\cot }^{-1}}\left( \tan x \right)dx} \\
 & \Rightarrow I=\int_{-2\pi }^{5\pi }{\left( \dfrac{\pi }{2}-x \right)dx} \\
\end{align}$
Now, we know that $\int_{a}^{b}{\left( f\left( x \right)\pm g\left( x \right) \right)dx=\int_{a}^{b}{f\left( x \right)}}dx\pm \int_{a}^{b}{g\left( x \right)dx}$. Using this property in I we get:
$\begin{align}
  & I=\int_{-2\pi }^{5\pi }{\left( \dfrac{\pi }{2}-x \right)dx} \\
 & \Rightarrow I=\int_{-2\pi }^{5\pi }{\dfrac{\pi }{2}dx}-\int_{-2\pi }^{5\pi }{xdx} \\
\end{align}$
Now, we know that $\int{k{{x}^{n}}dx}=k\int{{{x}^{n}}dx}$. Using this property in I we get:
$\begin{align}
  & I=\int_{-2\pi }^{5\pi }{\dfrac{\pi }{2}dx}-\int_{-2\pi }^{5\pi }{xdx} \\
 & \Rightarrow I=\dfrac{\pi }{2}\int_{-2\pi }^{5\pi }{dx}-\int_{-2\pi }^{5\pi }{xdx} \\
\end{align}$
Now, we also know that $\int_{a}^{b}{{{x}^{n}}dx}=\left[ \dfrac{{{x}^{n+1}}}{n+1} \right]_{a}^{b}=\left[ \dfrac{{{b}^{n+1}}}{n+1}-\dfrac{{{a}^{n+1}}}{n+1} \right]$
Thus, using this property in I we get:
\[\begin{align}
  & I=\dfrac{\pi }{2}\int_{-2\pi }^{5\pi }{dx}-\int_{-2\pi }^{5\pi }{xdx} \\
 & \Rightarrow I=\dfrac{\pi }{2}\int_{-2\pi }^{5\pi }{{{x}^{0}}dx}-\int_{-2\pi }^{5\pi }{xdx} \\
\end{align}\]
Now, applying the abovementioned property and applying the limits, we get:
\[\Rightarrow I=\dfrac{\pi }{2}\left[ \dfrac{{{x}^{0+1}}}{0+1} \right]_{-2\pi }^{5\pi }-\left[ \dfrac{{{x}^{1+1}}}{1+1} \right]_{-2\pi }^{5\pi }\]
Putting in the value of the limits we get:
\[\Rightarrow I=\dfrac{\pi }{2}\left[ \dfrac{{{\left( 5\pi \right)}^{1}}}{1}-\dfrac{{{\left( -2\pi \right)}^{1}}}{1} \right]-\left[ \dfrac{{{\left( 5\pi \right)}^{2}}}{2}-\dfrac{{{\left( -2\pi \right)}^{2}}}{2} \right]\]
Solving the value of these limits we get:
\[\begin{align}
  & \Rightarrow I=\dfrac{\pi }{2}\left[ 5\pi +2\pi \right]-\left[ \dfrac{25{{\pi }^{2}}-4{{\pi }^{2}}}{2} \right] \\
 & \Rightarrow I=\dfrac{7{{\pi }^{2}}}{2}-\dfrac{21{{\pi }^{2}}}{2} \\
 & \Rightarrow I=-\dfrac{14{{\pi }^{2}}}{2} \\
 & \Rightarrow I=-7{{\pi }^{2}} \\
\end{align}\]

Thus, the required value of I is $-7{{\pi }^{2}}$.

Note: The value of ${{\cot }^{-1}}\left( \tan x \right)$ obtained in equation (ii) can also be directly obtained by the following method:
We know that $\tan x=\cot \left( \dfrac{\pi }{2}-x \right)$
Thus, putting this value of $\tan x$ in ${{\cot }^{-1}}\left( \tan x \right)$ we get:
$\begin{align}
  & {{\cot }^{-1}}\left( \tan x \right) \\
 & \Rightarrow {{\cot }^{-1}}\left( \cot \left( \dfrac{\pi }{2}-x \right) \right) \\
\end{align}$
Now, we also know that ${{\cot }^{-1}}\left( \cot x \right)=x\text{ }\forall x\in R$
Thus, using this property in ${{\cot }^{-1}}\left( \tan x \right)$ we get:
$\begin{align}
  & \cot \left( \cot \left( \dfrac{\pi }{2}-x \right) \right) \\
 & \Rightarrow \dfrac{\pi }{2}-x \\
\end{align}$
The question can be further solved the same way as above.