<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] 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,
$I=\int_0^{\frac{\pi}{2}}cos(x)dx+4\times7+$
$\int_{15\pi-\frac{\pi}{2}}^{15\pi}(-cos(x))dx+\int_{15\pi}^{15\pi+\frac{\pi}{6}}(-cos(x))dx=1+28+1+\frac{1}{2}=\frac{61}{2}$
</s>