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Integration problem ?

  1. Oct 23, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]\int^{0}_{-\pi}\sqrt{1-cos^{2} x}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I substitute into (sin x)^2 and get an answer of -2 but the answer should be 2 . how to i do this question.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 23, 2007 #2

    cristo

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    How do you get an answer of -2? Perhaps you should show your work.
     
  4. Oct 23, 2007 #3
    just substitute (sin x)^2 and become sqr((sin x)^2) and then become sin x and substitute the limit from -pi to 0 get -2
     
  5. Oct 23, 2007 #4

    cristo

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    You need to integrate sin(x) before you plug in the limits.
     
  6. Oct 23, 2007 #5
    I have integrate it into -cos x and plugin the limit , i got -2 .But the answer is 2 . How ?
     
  7. Oct 23, 2007 #6

    cristo

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    So you have [tex]\left[-\cos(x)\right]^0_{-\pi}=cos(0)-cos(-\pi)[/tex]. Can you evaluate that?
     
  8. Oct 23, 2007 #7
    maybe you should be interpreting [itex]\sqrt{\sin^2(x)}[/itex] as [itex]|\sin(x)|[/itex]
     
  9. Oct 23, 2007 #8
     
    Last edited: Oct 23, 2007
  10. Oct 24, 2007 #9
    Then , how to integrate [itex]|\sin(x)|[/itex]
     
  11. Oct 24, 2007 #10

    HallsofIvy

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    For x between [itex]-\pi[/itex] and 0, sin(x)< 0. In that range, |sin(x)| is just -sin(x). Integrating that will obviously give you the negative of your previous answer.
     
  12. Oct 24, 2007 #11
    but ((sin x)^2 )^0.5 gives us two answers

    sinx and -sinx

    ???
     
  13. Oct 24, 2007 #12

    yeah, generally it does, but look here we are only integrating in [-pi.0], and obviously sinx, where x is from [-pi,0] is always negative, so

    I sin(x) I = -sinx, whenever x is from the interval [-pi. 0]
    now as halls said, integrating this you will get the desired answer.
     
  14. Oct 25, 2007 #13
    But i use mathematica to integrate , it shows -Cot x ((Sin x)^2)^(1/2)
    How to integrate to get this form ?
     
  15. Oct 25, 2007 #14

    HallsofIvy

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    Why would you want to? This is a definite integral. The result is a number. The integrand reduces to |sin(x)| which, for [itex]-\pi\le x\le 0[/itex] is -sin(x). That's easy to integrate.

    I've never used mathematica and what you give makes me glad I haven't! It's clearly using some general algorithm and then not recognizing that, since [itex](sin^2(x))^(1/2)[/itex] is |sin(x)|, [itex]-cot(x)(sin^2(x))^(1/2)= -(cos(x)/sin(x))(-sin(x))= cos(x)[/itex].
     
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