Answer
$$\int \tan ^{n} x d x =\frac{\tan^{n-1}x}{n-1}-\int \tan ^{n-2} x d x$$
Work Step by Step
Given $$\int \tan ^{n} x d x$$
Since
\begin{align*}
\int \tan ^{n} x d x&=\int \tan ^{n-2} x \tan ^{2} x d x\\
&=\int \tan ^{n-2} x\left(\sec ^{2} x-1\right) d x\\
&=\int \tan ^{n-2} x \sec ^{2} x d x-\int \tan ^{n-2} x d x
\end{align*}
Let $u=\tan x \ \Rightarrow \ \ du =\sec^2 x dx$, then
\begin{align*}
\int \tan ^{n-2} x \sec ^{2} x d x&=\int u ^{n-2} d u\\
&=\frac{u^{n-1}}{n-1}\\
&=\frac{\tan^{n-1}x}{n-1}+c
\end{align*}
Hence
$$\int \tan ^{n} x d x =\frac{\tan^{n-1}x}{n-1}-\int \tan ^{n-2} x d x$$
