Primitive of Power of Hyperbolic Sine of a x

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Theorem

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


Proof

With a view to expressing the problem in the form:

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

let:

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


and let:

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


Then:

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

$\blacksquare$


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