## Calculus: Early Transcendentals 8th Edition

= $\frac{xsinh(ax)}{a}$ + $\frac{cosh(ax)}{a^{2}}$+C
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