Question: I am trying to evaluate the exact value of the following definite integral:

\begin{align} \int^{\frac{91\pi}{6}}\_0 |\cos(x)| \, \mathrm{d}x \end{align}

Since $ \int^{\frac{\pi}{2}}\_0 \cos(x) \, \mathrm{d}x$ has symmetry, I did the following:
\begin{align}
\frac{91\pi}{6} \cdot \frac{2}{\pi} = \frac{91}{3}
\end{align}
\begin{align}
\int^{\frac{\pi}{2}}\_0 \cos(x) \, \mathrm{d}x = [\sin(x)]^{\frac{\pi}{2}}\_0 = 1
\end{align}
\begin{align}
\therefore \text{Bounded Area } = 1 \cdot \frac{91}{3} = \frac{91}{3}
\end{align}

Apparently, the correct answer is $ \frac{61}{2} $ which is very close to my answer. I cannot understand why my answer is wrong. Could someone please advise me?

Answer: 
Observe that

$$\int\_0^\pi|\cos x|dx=\left.2\int\_0^{\pi/2}\cos x dx=2\sin x\right|\_0^{\pi/2}=2\implies$$

$$\int\_0^{91\pi/6=15\pi+\frac\pi6}|\cos x|dx=\left.2\cdot15+\int\_\pi^{\frac{7\pi}6}(-\cos x)dx=30-\sin x\right|\_\pi^{\frac{7\pi}6}=30+\frac12=\frac{61}2$$
for $k\in \mathbb Z$

$\int\_{\frac{\pi}{2}+2k\pi}^{\frac{3\pi}{2}+2k\pi}(-cos(x))dx+$
$\int\_{\frac{3\pi}{2}+2k\pi}^{\frac{5\pi}{2}+2k\pi}cos(x)=4$

so,

$I=\int\_0^{\frac{\pi}{2}}cos(x)dx+4\times7+$

$\int\_{15\pi-\frac{\pi}{2}}^{15\pi}(-cos(x))dx+\int\_{15\pi}^{15\pi+\frac{\pi}{6}}(-cos(x))dx=1+28+1+\frac{1}{2}=\frac{61}{2}$