# 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?

• I can't understand your argument of how "cosine having symmetry" is connected with what you did... – DonAntonio Oct 16 '16 at 17:39
• @DonAntonio Cosine having symmetry means that I can just simply multiply the area by a certain number of times because the areas of the other regions are all similar to the first one. In this question, I can multiply the bounded area from $0$ to $\pi$ by 30. – LanceHAOH Oct 17 '16 at 5:15
• @La Ok....that's exactly what I did in my answer, I just didn't understand what you mean...and the symmetry is for $\;\int|\cos x|dx\;$ . – DonAntonio Oct 17 '16 at 9:12

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.,

$$\int_0^{\frac{47\pi}{3}}|\cos x|dx = 31\cdot1+\int_\frac{\pi}{2}^\frac{2\pi}{3}|\cos x|dx = 31 - \int_\frac{\pi}{2}^\frac{2\pi}{3}\cos x dx = 31 - \frac{\sqrt{3}}{2} + 1 = 32 - \frac{\sqrt{3}}{2}$$

• Integer multiple meaning that I can multiply my calculated area by any integer and it will give the correct result? E.g. $3\int^{\frac{\pi}{2}}_0dx = \int^{\frac{3\pi}{2}}_0dx$ – LanceHAOH Oct 16 '16 at 17:49
• In the particular case, yes, $\int_0^{k\pi/2} |\cos x|dx = k\int_0^{\pi/2} |\cos x|dx$ for any integer k, for the reason you outlined in your OP. It's also the case that $\int_0^{k\pi/2} dx = k\int_0^{\pi/2} dx$, but I don't think that's what you meant to ask. – user361424 Oct 16 '16 at 17:53
• Thanks! Indeed. I forgot to include the $|cos(x)|$ in my integral – LanceHAOH Oct 16 '16 at 17:55
• One caveat you should keep in mind if you encounter similar problems in the future. Since the period of $|\cos x|$ is in fact $\pi$, you only get to start the remainder from 0 because 30 is even. If the integer quotient were odd, the integral for the remainder would start from $\frac{\pi}{2}$. I'll add this to the answer with a worked example shortly. – user361424 Oct 17 '16 at 1:28
• In short I need to check where the graph of the remainder lies right? – LanceHAOH Oct 17 '16 at 2:09

Observe that

$$\int_0^\pi|\cos x|dx=\left.2\int_0^{\pi/2}\cos x dx=2\sin x\right|_0^{\pi/2}=2\implies$$

$$\int_0^{91\pi/6=15\pi+\frac\pi6}|\cos x|dx=\left.2\cdot15+\int_\pi^{\frac{7\pi}6}(-\cos x)dx=30-\sin x\right|_\pi^{\frac{7\pi}6}=30+\frac12=\frac{61}2$$

You can't do so, but you can $\frac{90\pi}{6}⋅\frac{2}{π}+\int_0^\frac{\pi}{6}|cosx|dx=30+1/2$

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,

$I=\int_0^{\frac{\pi}{2}}cos(x)dx+4\times7+$

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