Evaluate: $\displaystyle \int sinh^6 x\ cosh\ xdx$
$\displaystyle =\frac{1}{2}x^2\ cos(h) \sin(h^6)+C$
There seems to be a problem with your given answer.
Let $\displaystyle y = sinh(x)$, then $\displaystyle dy = cosh(x) dx$
So:
$\displaystyle \int sinh^6(x) \, cosh(x) dx = \int \, y^6 dy$ = $\displaystyle \frac{1}{7}y^7 + C$ = $\displaystyle \frac{1}{7}sinh^7(x) + C$.
-Dan
You need to be carefull where the power goes. There is a convention that $\displaystyle \sinh^n(x)$ actually means $\displaystyle [\sinh(x)]^n$ (even though this is an abuse of notation). If you move the power to where you have it you will need to add appropriate brackets to make the meaning clear.
So what the original means is:
$\displaystyle \int [\sinh (x)]^6 \cosh(x) $
RonL