# 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.

• Jan 23, 2018 at 2:49

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.

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \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}

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