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

Evaluate the value of the limit ∫−2π5πcot⁡−1(tan⁡x)dx\int_{-2\pi }^{5\pi }{{{\cot }^{-1}}\left( \tan x \right)dx}.

Explanation

Solution

We here need to find the value of the definite integral ∫−2π5πcot⁡−1(tan⁡x)dx\int_{-2\pi }^{5\pi }{{{\cot }^{-1}}\left( \tan x \right)dx}. We will here convert cot⁡−1(tan⁡x){{\cot }^{-1}}\left( \tan x \right) into tan⁡−1(tan⁡x){{\tan }^{-1}}\left( \tan x \right) by using the property cot⁡−1x+tan⁡−1x=π2{{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2} and then by using the property tan⁡−1(tan⁡x)=x{{\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 ∫abxndx=[xn+1n+1]ab=[bn+1n+1−an+1n+1]\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 ∫−2π5πcot⁡−1(tan⁡x)dx\int_{-2\pi }^{5\pi }{{{\cot }^{-1}}\left( \tan x \right)dx}. Let this be ‘I’.
Thus, we can say that:
I=∫−2π5πcot⁡−1(tan⁡x)dxI=\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(tan⁡x){{\cot }^{-1}}\left( \tan x \right) .
We will try to make it in the form of tan⁡−1(tan⁡x){{\tan }^{-1}}\left( \tan x \right) because of the property given by:
tan⁡−1(tan⁡x)=x ∀x∈R{{\tan }^{-1}}\left( \tan x \right)=x\text{ }\forall x\in R
Now, we know the property given as:
cot⁡−1x+tan⁡−1x=π2{{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}
Now, if we replace ‘x’ in the argument of these functions with tan⁡−1(tan⁡x){{\tan }^{-1}}\left( \tan x \right) , we will get the following result:
cot⁡−1(tan⁡x)+tan⁡−1(tan⁡x)=π2{{\cot }^{-1}}\left( \tan x \right)+{{\tan }^{-1}}\left( \tan x \right)=\dfrac{\pi }{2}
⇒cot⁡−1(tan⁡x)=π2−tan⁡−1(tan⁡x)\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(tan⁡x)=x{{\tan }^{-1}}\left( \tan x \right)=x. Thus, we can write cot⁡−1(tan⁡x){{\cot }^{-1}}\left( \tan x \right) as:
cot⁡−1(tan⁡x)=π2−tan⁡−1(tan⁡x) ⇒cot⁡−1(tan⁡x)=π2−x \begin{aligned} & {{\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{aligned}
Thus, we get that cot⁡−1(tan⁡x)=π2−x{{\cot }^{-1}}\left( \tan x \right)=\dfrac{\pi }{2}-x …..(ii)
Now, we can substitute the value of cot⁡−1(tan⁡x){{\cot }^{-1}}\left( \tan x \right) from equation (ii).
Substituting the value of cot⁡−1(tan⁡x){{\cot }^{-1}}\left( \tan x \right) from equation (ii) in I we get:
I=∫−2π5πcot⁡−1(tan⁡x)dx ⇒I=∫−2π5π(π2−x)dx \begin{aligned} & 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{aligned}
Now, we know that ∫ab(f(x)±g(x))dx=∫abf(x)dx±∫abg(x)dx\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:
I=∫−2π5π(π2−x)dx ⇒I=∫−2π5ππ2dx−∫−2π5πxdx \begin{aligned} & 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{aligned}
Now, we know that ∫kxndx=k∫xndx\int{k{{x}^{n}}dx}=k\int{{{x}^{n}}dx}. Using this property in I we get:
I=∫−2π5ππ2dx−∫−2π5πxdx ⇒I=π2∫−2π5πdx−∫−2π5πxdx \begin{aligned} & 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{aligned}
Now, we also know that ∫abxndx=[xn+1n+1]ab=[bn+1n+1−an+1n+1]\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:

& 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{aligned}$$ Now, applying the abovementioned property and applying the limits, we get: $$\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{aligned} & \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{aligned}$$ **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: $\begin{aligned} & {{\cot }^{-1}}\left( \tan x \right) \\\ & \Rightarrow {{\cot }^{-1}}\left( \cot \left( \dfrac{\pi }{2}-x \right) \right) \\\ \end{aligned}$ 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: $\begin{aligned} & \cot \left( \cot \left( \dfrac{\pi }{2}-x \right) \right) \\\ & \Rightarrow \dfrac{\pi }{2}-x \\\ \end{aligned}$ The question can be further solved the same way as above.