Primitive of Hyperbolic Sine of a x by Hyperbolic Cosine of a x/Proof 4

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Theorem

$\ds \int \sinh a x \cosh a x \rd x = \frac {\sinh^2 a x} {2 a} + C$


Proof

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:

\(\ds u\) \(=\) \(\ds \sinh a x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds a \cosh a x\) Derivative of $\sinh a x$


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \cosh a x\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {\sinh a x} a\) Primitive of $\cosh a x$


Then:

\(\ds \int \sinh a x \cosh a x \rd x\) \(=\) \(\ds \paren {\sinh a x} \paren {\frac {\sinh a x} a} - \int \paren {\frac {\sinh a x} a} \paren {a \cosh a x} \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {\sinh^2 a x} a - \int \sinh a x \cosh a x \rd x + C\) simplifying
\(\ds \leadsto \ \ \) \(\ds 2 \int \sinh a x \cosh a x \rd x\) \(=\) \(\ds \frac {\sinh^2 a x} a + C\) gathering terms
\(\ds \leadsto \ \ \) \(\ds \int \sinh a x \cosh a x \rd x\) \(=\) \(\ds \frac {\sinh^2 a x} {2 a} + C\) simplifying

$\blacksquare$