Integral of secant cubed

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One of the more challenging indefinite integrals of elementary calculus is

$\int \sec^3 x \, dx = \frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x| + C.$

[edit] Derivation

This antiderivative may be found by integration by parts, as follows:

$\int \sec^3 x \, dx = \int u\,dv$

where

$\begin{align} u &{}= \sec x, \\ dv &{}= \sec^2 x\,dx, \\ du &{}= \sec x \tan x\,dx, \\ v &{}= \tan x. \end{align}$

Then

$\begin{align} \int \sec^3 x \, dx &{}= \int u\,dv \\ &{}= uv - \int v\,du \\ &{} = \sec x \tan x - \int \sec x \tan^2 x\,dx \\ &{}= \sec x \tan x - \int \sec x\, (\sec^2 x - 1)\,dx \\ &{}= \sec x \tan x - \int \sec^3 x \, dx + \int \sec x\,dx. \end{align}$

Next we add $\scriptstyle{}\int\sec^3 x\,dx$ to both sides of the equality just derived:

$\begin{align} 2 \int \sec^3 x \, dx &{}= \sec x \tan x + \int \sec x\,dx \\ &{}= \sec x \tan x + \ln|\sec x + \tan x| + C. \end{align}$

Then divide both sides by 2:

$\int \sec^3 x \, dx = \frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x| + C.$

[edit] Higher odd powers of secant

Just as the integration by parts above reduced the integral of secant cubed to the integral of secant to the first power, so a similar process reduces the integral of higher odd powers of secant to lower ones. This is the Secant Reduction Formula, which follows the syntax:

$\int \sec^n{cx} \, dx = \frac{\sec^{n-2}{cx} \tan {cx}}{c(n-1)} \,+\, \frac{n-2}{n-1}\int \sec^{n-2}{cx} \, dx \qquad \mbox{ (for }n \ne 1\mbox{)}\,\!$

Alternatively:

$\int \sec^n{cx} \, dx = \frac{\sec^{n-1}{cx} \sin {cx}}{c(n-1)} \,+\, \frac{n-2}{n-1}\int \sec^{n-2}{cx} \, dx \qquad \mbox{ (for }n \ne 1\mbox{)}\,\!$