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..