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?

Your technique only works for integer multiples.  After that, you're implicitly assuming that the integral will scale linearly, and that's not the case.  So instead of $\frac{91}{3} \cdot 1$, it should be $$30\cdot 1 + \int_0^\frac{\pi}{6}|\cos x| dx = 30 + \sin\left(\frac{\pi}{6}\right) = \frac{61}{2}$$
A caveat - the period of $|\cos x|$ is in fact $\pi$, not $\frac{\pi}{2}$, so if the integer quotient had been odd (for instance, if the original integral had been from 0 to $\frac{47\pi}{3}$), you would need to start the integral for the remainder from $\frac{\pi}{2}$ (a region where $\cos x$ is negative), e.g.,
$$\int_0^{\frac{47\pi}{3}}|\cos x|dx = 31\cdot1+\int_\frac{\pi}{2}^\frac{2\pi}{3}|\cos x|dx = 31 - \int_\frac{\pi}{2}^\frac{2\pi}{3}\cos x dx = 31 - \frac{\sqrt{3}}{2} + 1 = 32 - \frac{\sqrt{3}}{2}$$
