Question

Integrate the function $ {\tan ^2}\left( {2x - 3} \right) $ .

Answer 1

Hint: To solve this question first we should know that $ {\tan ^2}\theta = {\sec ^2}\theta - 1 $ , and if we integrate it individually then we will get the result. We majorly use integration by substitution in such questions to get the required , and do not forget to put the value of original variable at the last.

Complete step-by-step answer:
Given, the function is $ {\tan ^2}\left( {2x - 3} \right) $ .
Let $ {\text{I}} = \int {{{\tan }^2}\left( {2x - 3} \right)} {\text{d}}x $
As, $ {\tan ^2}\theta = {\sec ^2}\theta - 1 $
Here, $ \theta = 2x - 3 $ .
So,
 $
\Rightarrow {\text{I}} = \int {{{\tan }^2}\left( {2x - 3} \right)} {\text{d}}x \\
\Rightarrow {\text{I}} = \int {\left( {{{\sec }^2}\left( {2x - 3} \right) - 1} \right)} {\text{d}}x \\
\Rightarrow {\text{I}} = \int {{{\sec }^2}} \left( {2x - 3} \right){\text{d}}x - \int {1 \cdot {\text{d}}x} \;
  $
Let, $ {{\text{I}}_1} = \int {{{\sec }^2}} \left( {2x - 3} \right){\text{d}}x $
So, $ {\text{I}} = {{\text{I}}_1} - \int {1 \cdot {\text{d}}x} $ ………………….(i)
 $ {{\text{I}}_1} = \int {{{\sec }^2}} \left( {2x - 3} \right){\text{d}}x $ ………….(ii)
Let, $ t = 2x - 3 $ .
Now, differentiate the above equation with respect to x.
\[
  \dfrac{{{\text{d}}t}}{{{\text{d}}x}} = 2 \\
  \dfrac{{dt}}{2} = {\text{d}}x \;
 \]
Now, substitute the above value in the equation (ii) and also substitute the value of 2x-3 in the (ii) equation.
 $
\Rightarrow {{\text{I}}_1} = \int {{{\sec }^2}} t\dfrac{{{\text{d}}t}}{2} \\
\Rightarrow {{\text{I}}_1} = \dfrac{1}{2}\int {{{\sec }^2}} t{\text{d}}t \;
  $
As we know that $ \int {{{\sec }^2}} x{\text{d}}x = \tan x $ , so using this concept simplifies the above equation.
 $ {{\text{I}}_1} = \dfrac{1}{2}\tan t + {C_1} $
Now, again substitute the value of t in the above equation.
 $ {{\text{I}}_1} = \dfrac{1}{2}\tan \left( {2x - 3} \right) + {C_1} $
Now, substitute the above value in the equation (i).
 $
\Rightarrow {\text{I}} = \dfrac{1}{2}\tan \left( {2x - 3} \right) + {C_1} - \int {1 \cdot {\text{d}}x} \\
\Rightarrow {\text{I}} = \dfrac{1}{2}\tan \left( {2x - 3} \right) + {C_1} - x + {C_2} \\
\Rightarrow {\text{I}} = \dfrac{1}{2}\tan \left( {2x - 3} \right) - x + C \;
  $
So, when we integrate the function $ {\tan ^2}\left( {2x - 3} \right) $ , we the result as $ \dfrac{1}{2}\tan \left( {2x - 3} \right) - x + C $ .
So, the correct answer is “ $ \dfrac{1}{2}\tan \left( {2x - 3} \right) - x + C $ .”.

Note: Integration of Maths is a means of combining or summing up the elements to find the whole. It is a reverse differentiation process, where we reduce the functions into components. This approach is used to find a large scale for the summation. Both integration and differentiation are basic components of calculus..