Find the integral of the function $\tan ^{4} x$.

(N/A) To find the integral $I = \int \tan ^{4} x \, dx$,we can simplify the integrand as follows:
$\tan ^{4} x = \tan ^{2} x \cdot \tan ^{2} x$
Using the identity $\tan ^{2} x = \sec ^{2} x - 1$,we get:
$\tan ^{4} x = (\sec ^{2} x - 1) \tan ^{2} x = \sec ^{2} x \tan ^{2} x - \tan ^{2} x$
Substituting $\tan ^{2} x = \sec ^{2} x - 1$ again:
$\tan ^{4} x = \sec ^{2} x \tan ^{2} x - (\sec ^{2} x - 1) = \sec ^{2} x \tan ^{2} x - \sec ^{2} x + 1$
Now,integrating term by term:
$\int \tan ^{4} x \, dx = \int \sec ^{2} x \tan ^{2} x \, dx - \int \sec ^{2} x \, dx + \int 1 \, dx$
For the first integral,let $u = \tan x$,then $du = \sec ^{2} x \, dx$. Thus,$\int \sec ^{2} x \tan ^{2} x \, dx = \int u^{2} \, du = \frac{u^{3}}{3} = \frac{\tan ^{3} x}{3}$.
Substituting this back:
$\int \tan ^{4} x \, dx = \frac{\tan ^{3} x}{3} - \tan x + x + C$,where $C$ is the constant of integration.