Answer
$\int {{{\tan }^n}x} dx = \frac{{{{\tan }^{n - 1}x}}}{{n - 1}} - \int {{{\tan }^{n - 2}}x} dx,{\text{ }}\left( {n \ne 1} \right)$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^n}x} dx \cr
& {\text{Split the integrand using the property of exponents }}{a^m}{a^n} = {a^{m + n}} \cr
& \int {{{\tan }^n}x} dx = \int {{{\tan }^{n - 2 + 2}}x} dx \cr
& \int {{{\tan }^n}x} dx = \int {{{\tan }^{n - 2}}x{{\tan }^2}x} dx \cr
& {\text{Apply the pythagorean identity ta}}{{\text{n}}^2}x = {\sec ^2}x - 1 \cr
& \int {{{\tan }^n}x} dx = \int {{{\tan }^{n - 2}}x\left( {{{\sec }^2}x - 1} \right)} dx \cr
& \int {{{\tan }^n}x} dx = \int {\left( {{{\tan }^{n - 2}}x{{\sec }^2}x} \right)} dx - \int {{{\tan }^{n - 2}}x} dx \cr
& {\text{Integrating}}{\text{, let }}u = \tan x,{\text{ }}du = {\sec ^2}xdx,{\text{ then}} \cr
& \int {{{\tan }^n}x} dx = \overbrace {\int {\left( {{{\tan }^{n - 2}}x{{\sec }^2}x} \right)} dx}^{\int {{u^{n - 2}}} du} - \int {{{\tan }^{n - 2}}x} dx \cr
& \int {{{\tan }^n}x} dx = \frac{{{{\tan }^{n - 2 + 1}x}}}{{n - 2 + 1}} - \int {{{\tan }^{n - 2}}x} dx \cr
& {\text{Simplifying}} \cr
& \int {{{\tan }^n}x} dx = \frac{{{{\tan }^{n - 1}x}}}{{n - 1}} - \int {{{\tan }^{n - 2}}x} dx,{\text{ }}\left( {n \ne 1} \right) \cr} $$