How do you integrate ${(\tan x)^2}$?

Hint:
As we know about the integral, it is used for finding the areas, volumes and for central points, etc. We use it for finding the integration of a, particularly given identity. It is denoted by the $\int {} $ sign; it is done with respect to a variable which can be $x,y,z$, etc.

Complete step by step solution:
Given that –
According to question integrate ${(\tan x)^2}$
Let – $I = {(\tan x)^2}$
Now we can write it as ${(\tan x)^2} = {\tan ^2}x$ because both are same in the trigonometry identity
We know that the three basic identity of trigonometry are First is ${\sin ^2}x + {\cos ^2}x = 1$ and second is ${\tan ^2}x + 1 = {\sec ^2}x$ and the third identity is the ${\cot ^2}x + 1 = \cos e{c^2}x$.
We will use the second identity which is the ${\tan ^2}x + 1 = {\sec ^2}x$ , now we will find value of ${\tan ^2}x = {\sec ^2}x - 1$ which we will put in the $I = {\tan ^2}x$ then we will integrate it.
Now we will integrate the $I$ with respect to $x$
$ = \int I $
$ = \int {{{\tan }^2}xdx} $
Now we will put the value of ${\tan ^2}x = {\sec ^2}x - 1$ in the above equation then we get
$ = \int {({{\sec }^2}x - 1)} dx$
Now we know that we can distribute integration on addition and subtraction so we will distribute it on negative
$ = \int {{{\sec }^2}xdx - \int {1dx} } $
Now we know that the integration of $\int {{{\sec }^2}} xdx = \tan x$ and $\int {1dx = x} $ now we will put these values in above equation then we get
$ = \tan x - x + c$
Where $c$ is the constant value which we get when we do integration of any identity

Therefore the integration of ${(\tan x)^2}$ is the $\tan x - x + c$ which is our required answer.

Note:
We can solve it directly by only using the second identity formula by just putting the value of ${\tan ^2}x = {\sec ^2}x - 1$ and then we get the same answer in the three or four-step if this question is in the objective type otherwise follow the above method.