Because
\begin{align}
\int \tan^{2n}(x)\sec^{3}(x) \, dx = {} & \int \tan^{2n-1}(x)(1+\tan^2(x)) \, d\sec(x) \\[10pt]
= {} &\int \tan^{2n-1}(x)\,d\sec(x) + \int \tan^{2n+1}(x) \, d\sec(x) \\[10pt]
= {} &\left[\sec(x)\tan^{2n-1}(x)-(2n-1)\int \tan^{2n-2}(x)\sec^3(x)\,dx \right] \\[10pt]
& {} + \left[\sec(x)\tan^{2n+1}(x)-(2n+1)\int\tan^{2n}(x)\sec^3(x) \, dx \right]
\end{align}
so
$$\begin{align*}
(2n+2)\int \tan^{2n}(x)\sec^3(x) \, dx &= \sec(x)\tan^{2n-1}(x) + \sec(x)\tan^{2n+1}(x) \\
& \quad - (2n-1)\int \tan^{2n-2}(x)\sec^3(x) \, dx \\
&= \sec^3(x)\tan^{2n-1}(x) - (2n-1)\int \tan^{2n-2}(x)\sec^3(x)dx
\end{align*}$$
$$\begin{align*}
\int \tan^{2n}(x)\sec^3(x) \, dx &= \frac1{2n+2}\sec^3(x)\tan^{2n-1}(x) - \frac{2n-1}{2n+2} \int \tan^{2n-2}(x)\sec^3(x) \, dx
\end{align*}$$
This is the recursive formula! I hope I made it clear, Please feed me back if anything doesn't make sense.