How do you integrate ${\left( {\tan x} \right)^4}$?

Hint: In order to evaluate the integral of the given function, first of all, we will reduce it to the standard form by a proper substitution. After completing the substitution part, we will integrate the functions using the standard formulas of integrals mentioned below.

Formula used:
$\Rightarrow$$\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C,n \ne - 1$
$\Rightarrow$$\int {{{\sec }^2}xdx = \tan x + C} $

Complete step-by-step answer:
First of all, let us take
$\Rightarrow$$I = \int {{{\tan }^4}xdx} $
Now, we will be substituting ${\tan ^4}x = {\tan ^2}x{\tan ^2}x$ , then the integral becomes
$\Rightarrow$$ = \int {{{\tan }^2}x.{{\tan }^2}xdx} $
Again, we will be using the trigonometric identity, which is ${\tan ^2}x = {\sec ^2}x - 1$ ,
Using the trigonometric identity, the integral becomes
$\Rightarrow$$ = \int {{{\tan }^2}x\left( {{{\sec }^2}x - 1} \right)dx} $
Simplifying the brackets inside the integral, we get
$\Rightarrow$$ = \int {\left( {{{\tan }^2}x{{\sec }^2}x - {{\tan }^2}x} \right)} dx$
Now, we separate the two integrals, in order to solve them separately, then the integrals are
$\Rightarrow$$ = \int {{{\tan }^2}x{{\sec }^2}xdx - \int {{{\tan }^2}xdx} } $
$\Rightarrow$$ = \int {{{\left( {\tan x} \right)}^2}{{\sec }^2}x} dx - \int {({{\sec }^2}x - 1)dx} $
Now, we consider the first integral as ${I_1}$ in order to solve this by substitution method,
$\Rightarrow$$ = {I_1} - \int {{{\sec }^2}xdx} + \int {dx} $ - - - - - - - - - - - $(1.)$
Now,
$\Rightarrow$${I_1} = \int {{{\left( {\tan x} \right)}^2}{{\sec }^2}xdx} $ ,
We write it like this because it will be easier to solve and use the substitution method. Also, we know that the differentiation of $\tan x$ is the rest of the function. This way the substitution methods are used.
Let us take $u = \tan x$
We differentiate on both sides, then it becomes
$\Rightarrow$$\dfrac{{du}}{{dx}} = {\sec ^2}x$
$ \Rightarrow du = {\sec ^2}xdx$
Now we substitute these values in the integral ${I_1}$ , then this becomes as
$\Rightarrow$${I_1} = \int {{u^2}du} $
We solve this integral by using the formula, mentioned above.
$\Rightarrow$$ = \dfrac{{{u^{2 + 1}}}}{{2 + 1}} + {C_1}$
$\Rightarrow$$ = \dfrac{{{u^3}}}{3} + {C_1}$ .
Now by substituting $u = \tan x$ , the value of ${I_1}$becomes as
$\Rightarrow$${I_1} = \dfrac{{{{\tan }^3}x}}{3} + {C_1}$
Now, we put the value of ${I_1}$ in $(1.)$,
$\Rightarrow$$I = \dfrac{{{{\tan }^3}x}}{3} + {C_1} - \int {{{\sec }^2}xdx} + \int {dx} $
Here again, we will be using the formula of integrals mentioned above, we get
$\Rightarrow$$ = \dfrac{{{{\tan }^3}x}}{3} - \tan x + x + C$ ,
where $C$ is all constants including ${C_1}$

∴$I = \dfrac{{{{\tan }^3}x}}{3} - \tan x + x + C$

Note:
The integral ${I_1}$ should be simplified separately, in order to use the substitution method properly. After obtaining the value of ${I_1}$, substitute it back in $I$ and evaluate the final solution of $I$. Here, $C$ is an arbitrary constant known as the constant of integration.