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?
 A: Observe that
$$\int_0^\pi|\cos x|dx=\left.2\int_0^{\pi/2}\cos x dx=2\sin x\right|_0^{\pi/2}=2\implies$$
$$\int_0^{91\pi/6=15\pi+\frac\pi6}|\cos x|dx=\left.2\cdot15+\int_\pi^{\frac{7\pi}6}(-\cos x)dx=30-\sin x\right|_\pi^{\frac{7\pi}6}=30+\frac12=\frac{61}2$$
A: 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.,
$$\int_0^{\frac{47\pi}{3}}|\cos x|dx = 31\cdot1+\int_\frac{\pi}{2}^\frac{2\pi}{3}|\cos x|dx = 31 - \int_\frac{\pi}{2}^\frac{2\pi}{3}\cos x dx = 31 - \frac{\sqrt{3}}{2} + 1 = 32 - \frac{\sqrt{3}}{2}$$
A: You can't do so, but you can $\frac{90\pi}{6}⋅\frac{2}{π}+\int_0^\frac{\pi}{6}|cosx|dx=30+1/2$
A: 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,
$I=\int_0^{\frac{\pi}{2}}cos(x)dx+4\times7+$
$\int_{15\pi-\frac{\pi}{2}}^{15\pi}(-cos(x))dx+\int_{15\pi}^{15\pi+\frac{\pi}{6}}(-cos(x))dx=1+28+1+\frac{1}{2}=\frac{61}{2}$
