Calculate the rotation area when $y = \cos(\frac{\pi x}{6})$, $\left| x \right| \leq 3$ rotates around the x-axis. Question 
Calculate the rotation area when $y = \cos(\frac{\pi x}{6})$, $\left| x \right| \leq 3$ rotates around the x-axis.
Attempt at solution
$$ A = 2\pi \int_{-3}^{3} f(x) \sqrt{1+ f'(x)^2} dx$$
Which in our case can be written as:
$$ A = 2\pi \int_{-3}^{3} \cos \left(\frac{\pi x}{6}\right) \sqrt{ 1 + \frac{1}{36} \pi^2 \sin^2\left(\frac{\pi x}{6}\right)} dx$$
By substitution $u = \sin \left(\frac{\pi x}{6}\right)$ and $du = \cos \left(\frac{\pi x}{6}\right)dx$, we get:
$$ A = 2\pi \int_{-1}^{1} \sqrt{ 1 + \frac{1}{36} \pi^2 u^2 }dx$$
Here I am stuck...

$$ A = 2\pi \int_{-3}^{3} \cos \left(\frac{\pi x}{6}\right) \sqrt{ 1 + \frac{1}{36} \pi^2 \sin^2\left(\frac{\pi x}{6}\right)} dx$$
By substitution $u = \sin \left(\dfrac{\pi x}{6}\right)$ and $du = \dfrac{\pi }{6}\cos \left(\dfrac{\pi x}{6}\right)dx$, we get:
$$ A = 2\pi\dfrac{6}{\pi} \int_{-1}^{1} \sqrt{ 1 + \frac{1}{36} \pi^2 u^2 }dx$$
let $\dfrac{1}{6}\pi u=\sinh t\,$:
$$A=\frac{72}{\pi } \int_{\sinh ^{-1}\left(-\frac{\pi }{6}\right)}^{\sinh ^{-1}\left(\frac{\pi }{6}\right)} \cosh ^2t \, dt$$
and with $2\cosh ^2t=1+\cosh2t$ we arrive this answer
$$ A = 2 \sqrt{\pi ^2+36}+\frac{72}{\pi } \sinh ^{-1}\left(\frac{\pi }{6}\right)$$
