Question

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

Answer 1

We are asked to integrate the given function. The angle of tangent is given as $\left( {2x - 3} \right)$ so for simplification assume the term $\left( {2x - 3} \right)$ as a whole to be a number. Then use the trigonometric identities to simplify the problem and then use the basic formulas for integration to solve the problem.

Complete step by step solution:
Given the function ${\tan ^2}\left( {2x - 3} \right)$
Now we integrate the function,
$I = \int {{{\tan }^2}\left( {2x - 3} \right)dx}$ (i)
Let $2x - 3 = t$ (ii)
Differentiating equation (ii), we get
$2dx = dt$
$\Rightarrow dx = \dfrac{1}{2}dt$ (iii)
Using equation (iii) and (ii) in (i) we get,
$I = \int {\left( {{{\tan }^2}t} \right)\left( {\dfrac{1}{2}dt} \right)}$
$\Rightarrow I = \dfrac{1}{2}\int {{{\tan }^2}tdt}$ (iv)
We have the trigonometric identity for tangent as,
$1 + {\tan ^2}\theta = {\sec ^2}\theta$
$\Rightarrow {\tan ^2}\theta = {\sec ^2}\theta - 1$
Therefore using this formula for ${\tan ^2}t$ we have,
${\tan ^2}t = {\sec ^2}t - 1$
Substituting this value of ${\tan ^2}t$ in equation (iv) we get,
$I = \dfrac{1}{2}\int {\left( {{{\sec }^2}t - 1} \right)dt}$
$\Rightarrow I = \dfrac{1}{2}\int {{{\sec }^2}tdt} - \dfrac{1}{2}\int {dt}$ (v)
Integration of ${\sec ^2}\theta$ is $\tan \theta$. Therefore using this in equation (v) we get,
$I = \dfrac{1}{2}\tan t - \dfrac{1}{2}t + c$ (vi)
$c$ is constant of integration
Now, substituting equation (ii) in (vi), we get
$I = \dfrac{1}{2}\tan \left( {2x - 3} \right) - \dfrac{1}{2}\left( {2x - 3} \right) + c$
$\Rightarrow I = \dfrac{1}{2}\tan \left( {2x - 3} \right) - x + \dfrac{3}{2} + c$ (vii)
The term $\left( {\dfrac{3}{2} + c} \right)$ can be considered as constant so, we can write equation (vii) as,
$I = \dfrac{1}{2}\tan \left( {2x - 3} \right) - x + C$
where $C$ is constant.
Therefore, integration of the function ${\tan ^2}\left( {2x - 3} \right)$ is $\dfrac{1}{2}\tan \left( {2x - 3}\right) - x + C$.

Note: Integration means adding small components to form a whole function. Here, we were asked to integrate a trigonometric function, to solve such types of questions, you will need to remember the integration of basic trigonometric functions that are sine, cosine and tangent. There are three other trigonometric functions which can be written in terms of the basic functions, these are cosecant which is inverse of sine, secant which is inverse of cosine and cotangent which is inverse of tangent. Also, while solving questions related to trigonometry, you should always remember the basic trigonometric identities.