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Find the integral from -π to x of |1/2 + cos t|

by RoRi

For 0 \leq x \leq \pi, compute the following integral:

    \[ \int_{-\pi}^x \left| \frac{1}{2} + \cos t \right| \, dt. \]

Since we are taking the integral of an absolute value, first we want to determine on which intervals \frac{1}{2} + \cos t is positive and on which it is negative. We have,

    \[ \left| \frac{1}{2} + \cos t \right| = \begin{dcases} \frac{1}{2} + \cos t & \text{if } -\pi \leq x \leq -\frac{2 \pi}{3}, \text{ or } \frac{2 \pi}{3} \leq x \leq \pi \\ - \left( \frac{1}{2} + \cos t \right) & \text{if } -\frac{2 \pi}{3} \leq x \leq \frac{2 \pi}{3} \end{dcases}. \]

Now must consider two cases. First, if 0 \leq x \leq \frac{2 \pi}{3}, then,

    \begin{align*} \int_{-\pi}^x \left| \frac{1}{2} + \cos t \right| \, dt &= - \int_{-\pi}^{- \frac{2 \pi}{3}} \left( \frac{1}{2} + \cos t \right) \, dt + \int_{- \frac{2 \pi}{3}}^x \left( \frac{1}{2} + \cos t \right) \, dt \\ &= - \left( - \frac{\pi}{3} + \frac{\pi}{2} \right) - \left( \sin \left(-\frac{2 \pi}{3} \right) - \sin (- \pi) \right) + \left( \frac{x}{2} + \frac{\pi}{3} \right) + \left( \sin x - \sin \left(-\frac{2 \pi}{3} \right) \right) \\ &= \frac{\pi}{6} + \sqrt{3} + \frac{x}{2} + \sin x. \end{align*}

In the other case, that \frac{2 \pi}{3} \leq x \leq \pi, then,

    \begin{align*} \int_{-\pi}^x \left| \frac{1}{2} + \cos t \right| \, dt &= -\int_{-\pi}^{-\frac{2 \pi}{3}} \left( \frac{1}{2} + \cos t \right) \, dt + \int_{- \frac{2 \pi}{3}}^{\frac{2 \pi}{3}} \left( \frac{1}{2} + \cos t \right) \,dt - \int_{\frac{2 \pi}{3}}^x \left( \frac{1}{2} + \cos t \right) \, dt \\ &= \left( - \frac{\pi}{6} + \frac{\sqrt{3}}{2} \right) + \left( \frac{2 \pi}{3} + \sqrt{3} \right) - \left( \frac{x}{2} - \frac{\pi}{3} \right) - \left( \sin x - \frac{\sqrt{3}}{2} \right) \\ &= \frac{5 \pi}{6} + 2 \sqrt{3} - \frac{x}{2} - \sin x. \end{align*}

4 comments

  1. eastlunder says:

    Also consider this solution:

    Using the additive property we have

    *** QuickLaTeX cannot compile formula:
\[
  \int_{-\pi}^{x}\Big|\frac{1}{2}+\cos{t}\Big|\, dt
  =
  \int_{-\pi}^{0}\Big|\frac{1}{2}+\cos{t}\Big|\, dt + \int_{0}^{x}\Big|\frac{1}{2}+\cos{t}\Big|\, dt,
\]

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    and applying Thm. 1.19 to the first integral we get

    *** QuickLaTeX cannot compile formula:
\[
  \int_{-\pi}^{0}\Big|\frac{1}{2}+\cos{t}\Big|\, dt + \int_{0}^{x}\Big|\frac{1}{2}+\cos{t}\Big|\, dt
  =
  -\int_{\pi}^{0}\Big|\frac{1}{2}+\cos{(-t)}\Big|\, dt + \int_{0}^{x}\Big|\frac{1}{2}+\cos{t}\Big|\, dt
\]

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    but \cos is an even function and 

    *** QuickLaTeX cannot compile formula:
\pi > 0

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    so Def. 1.4 yields

    *** QuickLaTeX cannot compile formula:
\begin{align*}
  -\int_{\pi}^{0}\Big|\frac{1}{2}+\cos{(-t)}\Big|\, dt + \int_{0}^{x}\Big|\frac{1}{2}+\cos{t}\Big|\, dt
  &=
    \underbrace{\int_{0}^{\pi}\Big|\frac{1}{2}+\cos{t}\Big|\, dt}_{\text{Prev. exercise.}} + \int_{0}^{x}\Big|\frac{1}{2}+\cos{t}\Big|\, dt=\\
  &= \Big(\frac{\pi}{6} + \sqrt{3}\Big) +\int_{0}^{x}\Big|\frac{1}{2}+\cos{t}\Big|\, dt.
\end{align*}

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    Again, from the previous exercise we know that the integrand has the property

    *** QuickLaTeX cannot compile formula:
\[
  \Big|\frac{1}{2}+\cos{t}\Big| =
  \begin{cases}
    \frac{1}{2}+\cos{t} & \text{if } -\frac{2\pi}{3} \leq t \leq \frac{2\pi}{3}\\
    -\big(\frac{1}{2}+\cos{t}\big) &  \text{if }  \frac{2\pi}{3} \leq t \leq \pi \text{ or } -\pi \leq t \leq -\frac{2\pi}{3}
  \end{cases}
\]

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    and since we now only have to solve an integral over [0,x] with 

    *** QuickLaTeX cannot compile formula:
x \in [0,\pi]

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    we need only worry about t \geq 0. However, because the integrand changes sign mid interval (depending on x) we must split into cases. Thus

    *** QuickLaTeX cannot compile formula:
\[
  \int_{0}^{x}\Big|\frac{1}{2}+\cos{t}\Big|\, dt =
  \begin{cases}
    \int_{0}^{x}\big[\frac{1}{2}+\cos{t}\big]\, dt & \text{if } x \in \big[0,\frac{2\pi}{3}\big]\\
    \int_{0}^{\frac{2\pi}{3}}\big[\frac{1}{2}+\cos{t}\big]\, dt - \int_{\frac{2\pi}{3}}^{x}\big[\frac{1}{2}+\cos{t}\big]\, dt & \text{if } x \in \big[\frac{2\pi}{3}, \pi\big]
  \end{cases}
\]

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    where the first case gives

    *** QuickLaTeX cannot compile formula:
\begin{align*}
  \int_{-\pi}^{x}\Big|\frac{1}{2}+\cos{t}\Big|\, dt &= \Big(\frac{\pi}{6} + \sqrt{3}\Big) +\int_{0}^{x}\Big|\frac{1}{2}+\cos{t}\Big|\, dt=\\
   &= \Big(\frac{\pi}{6} + \sqrt{3}\Big) + \int_{0}^{x}\Big[\frac{1}{2}+\cos{t}\Big]\, dt =\\
                                                                                     &= \Big(\frac{\pi}{6} + \sqrt{3}\Big) + \Big(\frac{x}{2}-0\Big) + (\sin{x}-\sin{0})= \\
  &=\frac{\pi}{6} + \sqrt{3} + \frac{x}{2} + \sin{x},
\end{align*}

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    and the second

    *** QuickLaTeX cannot compile formula:
\begin{align*}
  \int_{-\pi}^{x}\Big|\frac{1}{2}+\cos{t}\Big|\, dt &=\Big(\frac{\pi}{6} + \sqrt{3}\Big) +\int_{0}^{x}\Big|\frac{1}{2}+\cos{t}\Big|\, dt=\\
  &=\Big(\frac{\pi}{6} + \sqrt{3}\Big) + \int_{0}^{\frac{2\pi}{3}}\Big[\frac{1}{2}+\cos{t}\Big]\, dt - \int_{\frac{2\pi}{3}}^{x}\Big[\frac{1}{2}+\cos{t}\Big]\, dt=\\
                                                    &=\Big(\frac{\pi}{6} + \sqrt{3}\Big) + \Big(\frac{\pi}{3}+\sin \Big(\frac{2\pi}{3}\Big)\Big) - \Big(\frac{x}{2}-\frac{\pi}{3}+\sin x -\sin \Big(\frac{2\pi}{3}\Big)\Big)=\\
  &= \frac{5\pi}{6}+2\sqrt{3}-\frac{x}{2}-\sin x.
\end{align*}

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