Primitive of Cube of Secant of a x

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This article is complete as far as it goes, but it could do with expansion.
In particular: See Integral of secant cubed for multiple ways of solving this -- it is a classic student's exercise to solve this in as many ways as are available.
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Theorem

$\ds \int \sec^3 a x \rd x = \frac 1 {2 a} \paren {\sec a x \tan a x + \ln \size {\sec a x + \tan a x} } + C$


Proof 1

\(\ds \int \sec^3 x \rd x\) \(=\) \(\ds \frac {\sec a x \tan a x} {2 a} + \frac 1 2 \int \sec a x \rd x\) Primitive of $\sec^n a x$ where $n = 3$
\(\ds \) \(=\) \(\ds \frac {\sec a x \tan a x} {2 a} + \frac 1 2 \paren {\frac 1 a \ln \size {\sec a x + \tan a x} }\) Primitive of $\sec a x$
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \paren {\sec a x \tan a x + \ln \size {\sec a x + \tan a x} } + C\) simplifying

$\blacksquare$


Proof 2

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\ds u\) \(=\) \(\ds \sec a x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds a \sec a x \tan a x\) Derivative of Function of Constant Multiple, Derivative of $\sec$


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \sec^2 a x\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {\tan a x} a\) Primitive of $\sec^2 a x$


Then:

\(\ds \int \sec^3 a x \rd x\) \(=\) \(\ds \int \sec a x \sec^2 a x \rd x\)
\(\ds \) \(=\) \(\ds \sec a x \frac {\tan a x} a - \int \paren {\frac {\tan a x} a} \paren {a \sec a x \tan a x} \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac 1 a \sec a x \tan a x - \int \tan^2 a x \sec a x \rd x + C\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 a \sec a x \tan a x - \int \paren {\sec^2 a x - 1} \sec a x \rd x + C\) Difference of $\sec^2$ and $\tan^2$
\(\ds \) \(=\) \(\ds \frac 1 a \sec a x \tan a x - \int \sec^3 a x \rd x + \int \sec a x \rd x + C\) Linear Combination of Primitives
\(\ds \leadsto \ \ \) \(\ds 2 \int \sec^3 a x \rd x\) \(=\) \(\ds \frac 1 a \sec a x \tan a x + \int \sec a x \rd x + C\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 a \sec a x \tan a x + \frac 1 a \ln \size {\sec a x + \tan a x} + C\) Primitive of $\sec a x$

The result follows.

$\blacksquare$


Also see


Sources

  • 1967: Michael Spivak: Calculus: Part $\text {III}$: Derivatives and Integrals: Chapter $18$: Integration in Elementary Terms: Problems $3 \ \text {viii}$
This is a tricky and important integral that often comes up.
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