Primitive of x by Secant of a x

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Theorem

\(\ds \int x \sec a x \rd x\) \(=\) \(\ds \frac 1 {a^2} \paren {\frac {\paren {a x}^2} 2 - \frac {\paren {a x}^4} 8 + \frac {5 \paren {a x}^6} {144} - \cdots + \frac {\paren {-1}^n E_{2 n} \paren {a x}^{2 n + 2} } {\paren {2 n + 2} \paren {2 n}!} + \cdots} + C\)
\(\ds \) \(=\) \(\ds \frac 1 {a^2} \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n E_{2 n} \paren {a x}^{2 n + 2} } {\paren {2 n + 2} \paren {2 n}!} + C\)

where $E_{2 n}$ is the $2 n$th Euler number.


Proof

\(\ds \int x \sec a x \rd x\) \(=\) \(\ds \frac 1 {a^2} \int \theta \sec \theta \rd \theta\) Substitution of $a x \to \theta$
\(\ds \) \(=\) \(\ds \frac 1 {a^2} \int \theta \sum_{n \mathop = 0}^\infty \frac{ \paren {-1}^n E_{2 n} \theta^{2 n} } {\paren {2 n}!} \rd \theta\) Power Series Expansion for Secant Function
\(\ds \) \(=\) \(\ds \frac 1 {a^2} \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n E_{2 n} } {\paren {2 n}!} \int \theta^{2 n + 1} \rd \theta\) Power Series is Termwise Integrable within Radius of Convergence
\(\ds \) \(=\) \(\ds \frac 1 {a^2} \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n E_{2 n} \paren {a x}^{2 n + 2} } {\paren {2 n + 2} \paren {2 n}!} + C\) Substituting back $\theta \to ax$

$\blacksquare$


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