Question

 Find the value of the integral $\int {{{\left( {\tan x - \cot x} \right)}^2}dx} $.
\[
  {\text{A}}{\text{. }}\tan x + x + c \\
  {\text{B}}{\text{. }}\tan x - x + c \\
  {\text{C}}{\text{. }}\tan x - \cot x + c \\
  {\text{D}}{\text{. }}\tan x - \cot x - 4x + c \\
 \]

Answer 1

Hint: Here, we will proceed by using the formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ and then some basic formulas of trigonometry i.e., ${\left( {\sec x} \right)^2} = 1 + {\left( {\tan x} \right)^2}$ and ${\left( {\text{cosec x}} \right)^2} = 1 + {\left( {\cot x} \right)^2}$ are used in order to evaluate the value of the integral.

Complete Step-by-Step solution:
Let us suppose the integral ${\text{I}} = \int {{{\left( {\tan x - \cot x} \right)}^2}dx} $
Using the formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, the above integral becomes
$ \Rightarrow {\text{I}} = \int {\left[ {{{\left( {\tan x} \right)}^2} + {{\left( {\cot x} \right)}^2} - 2\left( {\tan x} \right)\left( {\cot x} \right)} \right]dx} $
Using the definition of cotangent trigonometric function i.e., $\cot x = \dfrac{1}{{\tan x}}$, we get
$
   \Rightarrow {\text{I}} = \int {\left[ {{{\left( {\tan x} \right)}^2} + {{\left( {\cot x} \right)}^2} - 2\left( {\tan x} \right)\left( {\dfrac{1}{{\tan x}}} \right)} \right]dx} \\
   \Rightarrow {\text{I}} = \int {\left[ {{{\left( {\tan x} \right)}^2} + {{\left( {\cot x} \right)}^2} - 2} \right]dx} {\text{ }} \to {\text{(1)}} \\
 $
As we know that ${\left( {\sec x} \right)^2} = 1 + {\left( {\tan x} \right)^2}$
$ \Rightarrow {\left( {\tan x} \right)^2} = {\left( {\sec x} \right)^2} - 1{\text{ }} \to {\text{(2)}}$
Also we know that ${\left( {\cos ecx} \right)^2} = 1 + {\left( {\cot x} \right)^2}$
$ \Rightarrow {\left( {\cot x} \right)^2} = {\left( {{\text{cosec}}x} \right)^2} - 1{\text{ }} \to {\text{(3)}}$
Using equation (2) and equation (3) in equation (1), we get
$
   \Rightarrow {\text{I}} = \int {\left[ {{{\left( {\sec x} \right)}^2} - 1 + {{\left( {{\text{cosec}}x} \right)}^2} - 1 - 2} \right]dx} \\
   \Rightarrow {\text{I}} = \int {\left[ {{{\left( {\sec x} \right)}^2} + {{\left( {{\text{cosec}}x} \right)}^2} - 4} \right]dx} \\
   \Rightarrow {\text{I}} = \int {{{\left( {\sec x} \right)}^2}dx} + \int {{{\left( {{\text{cosec}}x} \right)}^2}dx} - 4\int {dx} \\
   \Rightarrow {\text{I}} = \tan x - \cot x - 4x + c \\
 $
where c is the constant of integration.
Therefore, the value of the integral $\int {{{\left( {\tan x - \cot x} \right)}^2}dx} $ is equal to $\tan x - \cot x - 4x + c$.
Hence, option D is correct.

Note: In this particular problem, ${\text{I}} = \int {\left[ {{{\left( {\tan x} \right)}^2} + {{\left( {\cot x} \right)}^2} - 2} \right]dx} $ can only be solved by converting the tangent and cotangent trigonometric terms into secant and cosecant trigonometric terms. Since, the given integral is an indefinite integral (i.e., limits of integration are not there) that’s why we have used the constant of integration.