<s>[INST] Evaluate $ \int^{\frac{91\pi}{6}}_0 |\cos(x)| \, \mathrm{d}x$ 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?
 [/INST] Observe that
$$\int_0^\pi|\cos x|dx=\left.2\int_0^{\pi/2}\cos x dx=2\sin x\right|_0^{\pi/2}=2\implies$$
$$\int_0^{91\pi/6=15\pi+\frac\pi6}|\cos x|dx=\left.2\cdot15+\int_\pi^{\frac{7\pi}6}(-\cos x)dx=30-\sin x\right|_\pi^{\frac{7\pi}6}=30+\frac12=\frac{61}2$$
</s>