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Math Help - Trig Integration with Absolute Value

  1. #1
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    Trig Integration with Absolute Value

    Hi, I don't know where to start on this problem:

     \int_0^{\frac{\pi}{2}} |\sin{x} -\cos{x}| \ dx

    Any hints will be appreciated, thanks.
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  2. #2
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    Start by eliminating the absolute value. Do this by deducing all the places between the upper and lower bounds of integration where sin x - cos x is negative. Split the question up into separate integrals summed for each region, using multiplication by -1 to make the negative sections positive.

    Then integrate as per normal.
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  3. #3
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    \left| {\sin (t) - \cos (t)} \right| = \left\{ \begin{gathered}<br />
  \cos (t) - \sin (t),\,t \in \left[ {0,\pi /4} \right] \hfill \\<br />
  \sin (t) - \cos (t),\,t \in \left[ {\pi /4,\pi /2} \right] \hfill \\ <br />
\end{gathered}  \right.
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  4. #4
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    Hello,

    \sin \frac \pi 4=\cos \frac \pi 4=\frac{\sqrt{2}}{2}

    Before this point, cos decreases and sin increases.
    So before \frac \pi 4, cos > sin --> sin - cos < 0

    \int_{0}^{\frac \pi 2} |\sin(x)-\cos(x)| ~dx=\int_{0}^{\frac \pi 4} |\underbrace{\sin(x)-\cos(x)}_{\text{negative}}| ~dx+\int_{\frac \pi 4}^{\frac \pi 2} |\underbrace{\sin(x)-\cos(x)}_{\text{positive}}| ~dx

    Recall that |x|=\left\{\begin{array}{ll} -x \quad if~x<0 \\ x \quad if~x\ge 0 \end{array} \right.

    This gives \int_{0}^{\frac \pi 2} |\sin(x)-\cos(x)| ~dx=\int_{0}^{\frac \pi 4} -(\sin(x)-\cos(x)) ~dx+\int_{\frac \pi 4}^{\frac \pi 2} \sin(x)-\cos(x) ~dx
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  5. #5
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    Hello, MagicS06!

     \int_0^{\frac{\pi}{2}} |\sin{x} -\cos{x}| \ dx

    \text{Graph }\sin x \text{ and }\cos x\text{ on that interval.}
    Code:
            |
            * *   * *
            |   *   :
            | * : * :
            |*  :  *:
            |   :   : 
          - * - + - * - -
            0   π  π
            |

    On \left[0,\frac{\pi}{4}\right]\!:\;\;\sin x < \cos x

    . . The area is: . \int^{\frac{\pi}{4}}_0\bigg(\cos x - \sin x\bigg)\,dx



    On \left[\frac{\pi}{4},\:\frac{\pi}{2}\right]\!:\;\;\sin x > \cos x

    . . The area is: . \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \bigg(\sin x - \cos x\bigg)\,dx


    Go for it!

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