Primitive of Power of Hyperbolic Sine of a x

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Theorem

$\ds \int \sinh^n a x \rd x = \frac {\sinh^{n - 1} a x \cosh a x} {a n} - \frac {n - 1} n \int \sinh^{n - 2} a x \rd x$

for $n \ne 0$.


Proof

With a view to expressing the problem 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:

\(\ds u\) \(=\) \(\ds \sinh^{n - 1} a x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \paren {n - 1} a \sinh^{n - 2} a x \cosh a x\) Chain Rule for Derivatives, Derivative of $\sinh a x$, Derivative of Power


and let:

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


Then:

\(\ds \int \sinh^n a x \rd x\) \(=\) \(\ds \int \sinh^{n - 1} a x \sinh a x \rd x\)
\(\ds \) \(=\) \(\ds \sinh^{n - 1} a x \paren {\frac {\cosh a x} a}\) Integration by Parts
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \int \paren {\frac {\cosh a x} a} \paren {\paren {n - 1} a \sinh ^{n - 2} a x \cosh a x} \rd x\)
\(\ds \) \(=\) \(\ds \frac {\sinh^{n - 1} a x \cosh a x} a\) simplifying
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \int \paren {n - 1} \sinh^{n - 2} a x \cosh^2 a x \rd x\)
\(\ds \) \(=\) \(\ds \frac {\sinh^{n - 1} a x \cosh a x} a\) Difference of Squares of $\cosh$ and $\sinh$
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \int \paren {n - 1} \sinh^{n - 2} a x \paren {1 + \sinh^2 a x} \rd x\)
\(\ds \) \(=\) \(\ds \frac {\sinh^{n - 1} a x \cosh a x} a - \paren {n - 1} \int \sinh^{n - 2} a x \rd x\) Linear Combination of Primitives
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \paren {n - 1} \int \sinh^n a x \rd x\)
\(\ds \leadsto \ \ \) \(\ds n \int \sinh^n a x \rd x\) \(=\) \(\ds \frac {\sinh^{n - 1} a x \cosh a x} a - \paren {n - 1} \int \sinh^{n - 2} a x \rd x\) rearranging
\(\ds \leadsto \ \ \) \(\ds \int \sinh^n a x \rd x\) \(=\) \(\ds \frac {\sinh^{n - 1} a x \cosh a x} {a n} - \frac {n - 1} n \int \sinh^{n - 2} a x \rd x\) dividing both sides by $n$

We note that if $n = 0$, then $\dfrac {n - 1} n$ is undefined, rendering the derivation invalid.

$\blacksquare$


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