Given $n\in\Bbb N$, $n > 2$, find the reduction formula for: $$ \int \cosh^n x dx $$
I've derived an answer, but the signs do not match, so my question would be: where is the mistake in my calculation?
I'll use integration by parts: $$ u = \cosh^{n-1}x \\ du = (n-1)\cosh^{n-2}x\sinh x\ dx\\ dv = \cosh x\ dx\\ v = \sinh x $$
Thus the integral becomes: $$ \begin{align} J_n &= uv - \int vdu \\ &= \cosh^{n-1}x\sinh x - (n-1)\int \sinh^2x\cosh^{n-2}x\ dx\\ &= \cosh^{n-1}x\sinh x - (n-1)\left(\int(\cosh^2x-1)\cosh^{n-2}x\ dx\right)\\ &= \cosh^{n-1}x\sinh x - (n-1)(J_n - J_{n-2}) \iff \\ J_n &= \frac{\cosh^{n-1}x\sinh x + (n-1)J_{n-2}}{n} \end{align} $$
However, the answer section suggests that (note the minus sign): $$ J_n = \color{red}{-}\frac{\cosh^{n-1}x\sinh x}{n} + \frac{n-1}{n}J_{n-2} $$
I don't see why the first term is negative. Did I make a mistake somewhere or is it a typo in the book?
Thank you!
Your solution is completely correct! Definitely a typo in your book. Checked my own table of integrals and the one from Wikipedia; both agree with you. Well done!