Answer
= $\frac{xsinh(ax)}{a}$ + $\frac{cosh(ax)}{a^{2}}$+C
Work Step by Step
Note: cosh(ax) is a hyperbolic function; for this question, we can treat it as if we were dealing with cos(x). Treat a as a constant for this problem.
Let u=x
u' = 1
$\frac{du}{dx}$ = 1
du = 1dx
du = dx
dv = cosh(ax)dx
$\int dv$ = $\frac{1}{a}$ $\int acosh(ax)dx$ [Chain Rule]
v = $\frac{1}{a}$ sinh(ax)
uv - $\int vdu$
x[$\frac{1}{a}$sinh(ax)] - $\frac{1}{a}$$\int sinh(ax)$dx
Let us solve the integral first before putting the entire answer together.
$\frac{1}{a}$$\int sinh(ax)$
Let y = ax
y' = a
$\frac{dy}{dx}$ = a
dy = adx
dx = $\frac{1}{a}$dy
Plug in all the values:
($\frac{1}{a})^{2}$$\int sinh(y)$dy
= -($\frac{1}{a})^{2}$cosh(ax) + C
Now combine everything:
x[$\frac{1}{a}$sinh(ax)] - $\frac{1}{a}$$\int sinh(ax)$dx
= x[$\frac{1}{a}$sinh(ax)] + ($\frac{1}{a})^{2}$cosh(ax) + C
