# Primitive of Power of Hyperbolic Sine of a x

## 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 $n$

$\blacksquare$