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Math Help - Incorrect integral

  1. #1
    Junior Member
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    Incorrect integral

    Hi, I was doing a question with this simple integral

    3a\int^{\frac{\pi}{2}}_0cosxsinxdx

    so rather than do it by parts (which probably would turn out to be double back on itself so you get the integral again), I opted for using sin2x = 2sinxcosx

    so that means the integral simplified to

    3a\int^{\frac{\pi}{2}}_0\frac{1}{2}sin2x

    which integrates to

    3a[-\frac{1}{4}cos2x]^{\frac{\pi}{2}}_0

    so wack in the limits and I get 3a[1-1] = 0

    but that isn't right is it?

    The answer scheme suggest 3a\int^{\frac{\pi}{2}}_0cosxsinxdx = 3a[\frac{1}{2}sin^{2}2x]^{\frac{\pi}{2}}_0

    but i) How come my method doesn't work
    ii) How can the mark scheme do that. You can't just do that!?

    Thanks
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by thomas49th View Post

    3a[-\frac{1}{4}cos2x]^{\frac{\pi}{2}}_0

    so wack in the limits and I get 3a[1-1] = 0



    3a[-\frac{1}{4}cos2x]^{\frac{\pi}{2}}_0

    =-\frac{3a}{4}[cos2x]^{\frac{\pi}{2}}_0

    =-\frac{3a}{4}[cos{\pi}-cos(0)]

    =-\frac{3a}{4}[-1-1]

    =\frac{3a}{2}
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  3. #3
    Super Member malaygoel's Avatar
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    Quote Originally Posted by thomas49th View Post

    The answer scheme suggest 3a\int^{\frac{\pi}{2}}_0cosxsinxdx = 3a[\frac{1}{2}sin^{2}2x]^{\frac{\pi}{2}}_0

    ii) How can the mark scheme do that. You can't just do that!?

    Thanks
    It is incorrect

    it should be
    3a[\frac{1}{2}sin^{2}x]^{\frac{\pi}{2}}_0
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  4. #4
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    im an idiot...

    how did they get there stupid <br /> <br />
3a[\frac{1}{2}sin^{2}2x]^{\frac{\pi}{2}}_0<br />

    thanks
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  5. #5
    MHF Contributor

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    The simplest way to do \int sin(x)cos(x) dx is neither "by parts" or using a trig identity like that. Let u= sin(x) so that du= cos(x) dx and this becomes the simple integral \int u du.

    An integral of that is \frac{1}{2} u^2. That's where they get 3a\left[\frac{1}{2}sin^2(x)\right]_0^{\pi/2}.
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  6. #6
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    oh course the one of the products is the differential of the other. cheers
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