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Thread: Integral of Abs Value Trigonometric function

  1. #1
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    Integral of Abs Value Trigonometric function

    So I have to integrate this function... looked simple enough at first but soon I found it was not so. I've found the answer using wolframalpha and solutions etc. but don't understand completely how to get there



    I first tried integrating sin(abs(x)) just to -cos(abs(x)) and got 0+1/sqrt(2), obviously wrong. Then I tried to break it up - integral of sinx=-cosx for the positive value and integral of sin(-x)=cosx for the negative value, but this just gave me -1/sqrt(2)...

    How exactly do I have to break this up to get it to work?

    Thank you for any advice!
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  2. #2
    Super Member Random Variable's Avatar
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    EDIT: No, that's wrong. Hold on a sec.

     \displaystyle = \int^{\pi /4}_{0} \sin x \ dx + \int^{\pi /2}_{0} \sin x \ dx
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    Quote Originally Posted by Random Variable View Post
    EDIT: No, that's wrong. Hold on a sec.
    Alright, thank you!
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    Quote Originally Posted by metaname90 View Post
    Alright, thank you!
    You need to break the original integral into two integrals 1 from -pi/4 to 0 and 0 to pi/2 since abs value is  -x, \ x<0 and x, \ x>0
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    Quote Originally Posted by dwsmith View Post
    You need to break into two integrals 1 from -pi/4 to 0 and 0 to pi/2
    That's what Random Variable said before saying that was incorrect... he had:

    - \int^{0}_{-\pi/4} \sin x \ dx + \int^{\pi/2}_{0} \sin x \ dx

    But I was about to ask some more questions about exactly how that applies to other cases and not just this case... was it that \int^{0}_{-\pi/4} \sin (-x) \ dx = \int^{0}_{-\pi/4} -\sin x \ dx = - \int^{0}_{-\pi/4} \sin x \ dx or what...
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    Quote Originally Posted by Random Variable View Post
    EDIT: No, that's wrong. Hold on a sec.

     \displaystyle = \int^{\pi /4}_{0} \sin x \ dx + \int^{\pi /2}_{0} \sin x \ dx
    Okay thank you, I see where you got that function but I'm still trying to understand this exactly... for the second integral everything is just normal because x > 0, and for the first one since x < 0 we need the integral of sin(-x) which equals -sin(x) then we can take - integral of sinx but when we flip the integral that becomes positive... so how does -pi/4 go to pi/4?

    Thanks!
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    Quote Originally Posted by metaname90 View Post
    That's what Random Variable said before saying that was incorrect... he had:

    - \int^{0}_{-\pi/4} \sin x \ dx + \int^{\pi/2}_{0} \sin x \ dx

    But I was about to ask some more questions about exactly how that applies to other cases and not just this case... was it that \int^{0}_{-\pi/4} \sin (-x) \ dx = \int^{0}_{-\pi/4} -\sin x \ dx = - \int^{0}_{-\pi/4} \sin x \ dx or what...
    You can do -Pi/4 to 0 too.
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  8. #8
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    Quote Originally Posted by metaname90 View Post
    Okay thank you, I see where you got that function but I'm still trying to understand this exactly... for the second integral everything is just normal because x > 0, and for the first one since x < 0 we need the integral of sin(-x) which equals -sin(x) then we can take - integral of sinx but when we flip the integral that becomes positive... so how does -pi/4 go to pi/4?

    Thanks!
    \displaystyle \int^{\pi /4}_{0} \sin x \ dx + \int^{\pi /2}_{0} \sin x \ dx \equiv -\int^{0}_{-\pi/4} \sin x \ dx + \int^{\pi /2}_{0} \sin x \ dx
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  9. #9
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    Quote Originally Posted by metaname90 View Post
    So I have to integrate this function... looked simple enough at first but soon I found it was not so. I've found the answer using wolframalpha and solutions etc. but don't understand completely how to get there



    I first tried integrating sin(abs(x)) just to -cos(abs(x)) and got 0+1/sqrt(2), obviously wrong. Then I tried to break it up - integral of sinx=-cosx for the positive value and integral of sin(-x)=cosx for the negative value, but this just gave me -1/sqrt(2)...

    How exactly do I have to break this up to get it to work?

    Thank you for any advice!
    Definition of |x|:

    |x|=\left\{\begin{array}{cc}\ x,&\mbox{ if }<br />
x>0\\ \ 0, &\mbox{ if }<br />
x=0\\ \ -x, & \mbox{ if } x<0\end{array}\right.

    sin(x) is an odd function, therefore:

    \sin\left(|x|\right)=\left\{\begin{array}{cc}\ \sin(x),&\mbox{ if }<br />
x>0\\ \ \sin(0), &\mbox{ if }<br />
x=0\\ \ -\sin(x), & \mbox{ if } x<0\end{array}\right.
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