integral of secant cubed

This antiderivative may be found by integration by parts, as follows:{{Cite book|last=Stewart|first=James|authorlink=James Stewart (mathematician)|title=Calculus - Early Transcendentals|publisher=Cengage Learning|year=2012|isbn=978-0-538-49790-9|location=United States|pages=475–6|chapter=Section 7.2: Trigonometric Integrals}}

:$\backslash int\; \backslash sec^3\; x\; \backslash ,\; dx\; =\; \backslash int\; u\backslash ,dv\; =\; uv\; -\; \backslash int\; v\; \backslash ,du$

where

:$$

u = \sec x,\quad dv = \sec^2 x\,dx,\quad v

= \tan x,\quad du = \sec x \tan x\,dx.

Then

:$\backslash begin\{align\}$

\int \sec^3 x \, dx

&= \int (\sec x)(\sec^2 x)\,dx \\

&= \sec x \tan x - \int \tan x\,(\sec x \tan x)\,dx \\

&= \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 - \left(\int \sec^3 x \, dx - \int \sec x\,dx\right) \\

&= \sec x \tan x - \int \sec^3 x \, dx + \int \sec x\,dx.

\end{align}

Next add $\backslash int\backslash sec^3\; x\; \backslash ,dx$ to both sides:{{efn|The constants of integration are absorbed in the remaining integral term.}}

:$\backslash begin\{align\}$

2 \int \sec^3 x \, dx

&= \sec x \tan x + \int \sec x\,dx \\

&= \sec x \tan x + \ln\left|\sec x + \tan x\right| + C,

\end{align}

using the integral of the secant function, $\backslash int\; \backslash sec\; x\; \backslash ,dx\; =\; \backslash ln\; \backslash left|\backslash sec\; x\; +\; \backslash tan\; x\backslash right|\; +\; C.$

Finally, divide both sides by 2:

:$$

\int \sec^3 x \, dx

= \tfrac12(\sec x \tan x + \ln \left|\sec x + \tan x\right|) + C,

which was to be derived. A possible mnemonic is: "The integral of secant cubed is the average of the derivative and integral of secant".

:$$

\int \sec^3 x \, dx

= \int \frac{dx}{\cos^3 x}

= \int \frac{\cos x\,dx}{\cos^4 x}

= \int \frac{\cos x\,dx}{(1-\sin^2 x)^2}

= \int \frac{du}{(1-u^2)^2}

where $u\; =\; \backslash sin\; x$, so that $du\; =\; \backslash cos\; x\backslash ,dx$. This admits a decomposition by partial fractions:

:$$

\frac{1}{(1-u^2)^2}

= \frac{1}{(1+u)^2(1-u)^2}

= \frac{1}{4(1+u)} + \frac{1}{4(1+u)^2} + \frac{1}{4(1-u)} + \frac{1}{4(1-u)^2}.

Antidifferentiating term-by-term, one gets

:$\backslash begin\{align\}$

\int \sec^3 x \, dx

&= \tfrac14 \ln |1+u| - \frac{1}{4(1+u)} - \tfrac14 \ln|1-u| + \frac{1}{4(1-u)} + C \\[6pt]

&= \tfrac14 \ln \Biggl| \frac{1+u}{1-u} \Biggl| + \frac{u}{2(1-u^2)} + C \\[6pt]

&= \tfrac14 \ln \Biggl|\frac{1+\sin x}{1-\sin x} \Biggl| + \frac{\sin x}{2\cos^2 x} + C\\[6pt]

&= \tfrac14 \ln \left|\frac{1+\sin x}{1-\sin x}\right| + \tfrac12 \sec x \tan x + C \\[6pt]

&= \tfrac14 \ln \left|\frac{(1+\sin x)^2}{1-\sin^2 x}\right| + \tfrac12 \sec x \tan x + C \\[6pt]

&= \tfrac14 \ln \left|\frac{(1+\sin x)^2}{\cos^2 x}\right| + \tfrac12 \sec x \tan x + C \\[6pt]

&= \tfrac12 \ln \left|\frac{1+\sin x}{\cos x}\right| + \tfrac12 \sec x \tan x + C \\[6pt]

&= \tfrac12 (\ln|\sec x + \tan x| + \sec x \tan x) + C.

\end{align}

Alternatively, one may use the tangent half-angle substitution for any rational function of trigonometric functions; for this particular integrand, that method leads to the integration of

:$$

\frac{2(1+u^2)^2}{(1-u^2)^3}

= \frac{1}{2(1+u)} - \frac{1}{2(1+u)^2} + \frac{1}{(1+u)^3} + \frac{1}{2(1-u)} - \frac{1}{2(1-u)^2} + \frac{1}{(1-u)^3}.

Integrals of the form: $\backslash int\; \backslash sec^n\; x\; \backslash tan^m\; x\backslash ,\; dx$ can be reduced using the Pythagorean identity if $n$ is even or $n$ and $m$ are both odd. If $n$ is odd and $m$ is even, hyperbolic substitutions can be used to replace the nested integration by parts with hyperbolic power-reducing formulas.

:$\backslash begin\{align\}$

\sec x &= \cosh u \\[6pt]

\tan x &= \sinh u \\[6pt]

\sec^2 x \, dx &= \cosh u \, du \text{ or } \sec x \tan x\, dx = \sinh u \, du \\[6pt]

\sec x \, dx &= \, du \text{ or } dx = \operatorname{sech} u \, du \\[6pt]

u &= \operatorname{arcosh} (\sec x ) = \operatorname{arsinh} ( \tan x ) = \ln|\sec x + \tan x|

\end{align}

Note that $\backslash int\; \backslash sec\; x\; \backslash ,\; dx\; =\; \backslash ln|\backslash sec\; x\; +\; \backslash tan\; x|$ follows directly from this substitution.

:$\backslash begin\{align\}$

\int \sec^3 x \, dx

&= \int \cosh^2 u\,du \\[6pt]

&= \tfrac12 \int ( \cosh 2u +1) \,du \\[6pt]

&= \tfrac12 \left( \tfrac12 \sinh2u + u\right) + C\\[6pt]

&= \tfrac12 ( \sinh u \cosh u + u ) + C \\[6pt]

&= \tfrac12 (\sec x \tan x + \ln \left|\sec x + \tan x\right|) + C

\end{align}

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 x \, dx

= \frac{\sec^{n-2} x \tan x}{n-1} \,+\, \frac{n-2}{n-1}\int \sec^{n-2} x \, dx \qquad \text{ (for }n \ne 1\text{)}\,\!

Even powers of tangents can be accommodated by using binomial expansion to form an odd polynomial of secant and using these formulae on the largest term and combining like terms.

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Category:Integral calculus