Find a reduction formula for a trigonometric integral $$\int\tan^{2n}(x)\sec^3(x)\,dx$$
I am aware that we use integration by parts to find the reduction formula, but I'm stuggling with what to use for $u,$ and $v.$
I've tried letting
$$u=\tan^{2n-2}(x)$$
I'm then left with,
$$dv=\sec^3(x)\tan^2(x) \, dx$$
which is fine, except the integral of $dv$ is very complicated.
Any help is much appreciated, thanks.
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
I_{nr} & \equiv \int\tan^{2n}\pars{x}\sec^{2r + 1}\pars{x}\,\dd x =
{1 \over 2r + 1}\int\tan^{2n - 1}\pars{x}\,\dd\sec^{2r + 1}\pars{x}
\\[1cm] & =
{\tan^{2n - 1}\pars{x}\sec^{2r + 1}\pars{x} \over 2r + 1} \\[2mm] &\ 
-\,{1 \over 2r + 1}\int\sec^{2r + 1}\pars{x}\pars{2n - 1}\tan^{2n - 2}\pars{x}\sec^{2}\pars{x}\,\dd x
\\[1cm] & =
{\tan^{2n - 1}\pars{x}\sec^{2r + 1}\pars{x} \over 2r + 1} -
{2n - 1 \over 2r + 1}\int
\tan^{2n - 2}\pars{x}\sec^{2r + 3}\pars{x}\,\dd x
\end{align}

\begin{equation}
\bbx{I_{nr} =
{\tan^{2n - 1}\pars{x}\sec^{2r + 1}\pars{x} \over 2r + 1} -
{2n - 1 \over 2r + 1}\,I_{n - 1,r + 1}}\label{1}\tag{1}
\end{equation}
which is equivalent to
\begin{equation}
I_{n + 1,r - 1} =
{\tan^{2n + 1}\pars{x}\sec^{2r - 1}\pars{x} \over 2r - 1} -
{2n + 1 \over 2r - 1}\,I_{nr}
\end{equation}
\begin{equation}
\implies
\bbx{I_{nr} =
{\tan^{2n + 1}\pars{x}\sec^{2r - 1}\pars{x} \over 2n + 1} -
{2r - 1 \over 2n + 1}\,I_{n + 1,r - 1}}
\label{2}\tag{2}
\end{equation}
