$($a$)$: Derive the following alternate form for the indefinite integral in identitiy $(6\mathrm{.}1\mathrm{.}5)($note: the simplest way to do this is to rewrite the right side of $(6\mathrm{.}1\mathrm{.}5)$ with trig identities and algebra$)$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}sin(ax)cos(bx)dx=\frac{bsin(ax)sin(bx)+acos(ax)cos(bx)}{-a^2+b^2},a+-b$

$($b$)$: Derive a similar identity for the integral

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}sinh(ax)cos(bx)dx,a*b0$

Write your answer in a form that just uses trig and hyperbolic trig functions, but no extra exponential terms $($i$.$e$.,$ no $e^{ax}$ should be explicit present, but having $cosh(ax)$ or the like is fine.