Answer
$$\int {{{\tan }^n}xdx} = \frac{1}{{n - 1}}{\tan ^{n - 1}}x - \int {{{\tan }^{n - 2}}xdx} $$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^n}xdx} \cr
& {\text{Recall that }}{a^{m + n}} = {a^m}{a^n},{\text{ then}} \cr
& \int {{{\tan }^n}xdx} = \int {{{\tan }^{n - 2 + 2}}xdx} \cr
& \int {{{\tan }^n}xdx} = \int {{{\tan }^{n - 2}}x{{\tan }^2}xdx} \cr
& {\text{Rewrite }}{\tan ^2}x{\text{ using the identity ta}}{{\text{n}}^2}x = {\sec ^2}x - 1 \cr
& \int {{{\tan }^n}xdx} = \int {{{\tan }^{n - 2}}x\left( {{{\sec }^2}x - 1} \right)dx} \cr
& \int {{{\tan }^n}xdx} = \int {{{\tan }^{n - 2}}x{{\sec }^2}xdx} - \int {{{\tan }^{n - 2}}xdx} \cr
& {\text{Integrate }}\int {{{\tan }^{n - 2}}x{{\sec }^2}xdx} {\text{ by using the power rule}} \cr
& \int {{{\tan }^n}xdx} = \frac{{{{\tan }^{n - 2 + 1}}x}}{{n - 2 + 1}} - \int {{{\tan }^{n - 2}}xdx} \cr
& {\text{Simplifying}} \cr
& \int {{{\tan }^n}xdx} = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - \int {{{\tan }^{n - 2}}xdx} \cr
& \int {{{\tan }^n}xdx} = \frac{1}{{n - 1}}{\tan ^{n - 1}}x - \int {{{\tan }^{n - 2}}xdx} \cr} $$