2
$\begingroup$

$$\int\tan^{2n}(x)\sec^3(x)\,dx$$ 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.

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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


\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}

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.