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Math Help - definite integrate

  1. #1
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    definite integrate

    evaluate \int^{\frac{\pi}{3}}_0(sinx-cosx)^2 dx



    what i did


    \int^{\frac{\pi}{3}}_0(sinx-cosx)^2 dx=[\frac{(sinx-cosx)^3}{3(-cosx-sinx)}]^{\frac{\pi}{3}}_0

    is this right?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    I think you can treat it as:

    \int (f(x))^n dx = \frac{f^{(n+1)}(x)}{(n+1)(f'(x))}

    EDIT: After you edited your post, yes, that's what I would have done
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  3. #3
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    i posted what i did using the formula u posted, mind to check if i did it right
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Hm... if y = sin(x)

    Then dy/dx becomes cos(x)

    And if y = cos(x),

    Then dy/dx = -sin(x)

    What I mean is that you put the differential of the function and not the integral of the function in the denominator.
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  5. #5
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    the 2 posts u posted are contradicting.. so am i right?
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Sorry, I have edited my first post without actually taking a look at whether the integral was right or wrong.

    My second post says it.

    In fact, it's only my edit in my first post which is contradicting. The 'formula' that I posted remains the same and applies.

    So, your integral is wrong. You need to put the differential of the function as denominator, not its integral.


    If you had to differentiate

    y = (x^2 + x + 1)^3

    This becomes:

    \frac{dy}{dx} = (x^2 + x + 1)^2 . (2x + 1)

    When you integrate, you divide instead of multiplying the differential of the function.
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  7. #7
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    i dont know what went wrong but substituting the values in doesnt give me the correct answer...would u work it out and tell me what u get?
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  8. #8
    MHF Contributor Unknown008's Avatar
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    \int^{\frac{\pi}{3}}_0(sinx-cosx)^2 dx = [\frac{(sinx - cosx)^3}{3(cosx + sinx)}]^{\frac{\pi}{3}} _0

    =  \frac{(sin(\frac{\pi}{3}) - cos(\frac{\pi}{3}))^3}{3(cos(\frac{\pi}{3}) + sin(\frac{\pi}{3}))} -  \frac{(sin(0) - cos(0))^3}{3(cos(0) + sin(0))}

    =  \frac{(\frac{\sqrt{3}}{2} - \frac{1}{2})^3}{3(\frac12 + \frac{\sqrt{3}}{2})} -  \frac{(0 - 1)^3}{3(1 + 0)}

    =  \frac{(\frac{\sqrt{3}-1}{2})^3}{3(\frac{1+\sqrt{3}}{2})} -  \frac{-1}{3}

    =  \frac{(\frac{(\sqrt{3}-1)^3}{8})}{\frac{3+3\sqrt{3}}{2}} +  \frac{1}{3}

    = \frac{6\sqrt{3}-10}{12 + 12\sqrt{3}} +  \frac{1}{3}

    If I didn't make any mistake, you can now further simplify.

    EDIT: Seems like it's wrong.
    Last edited by Unknown008; September 8th 2010 at 08:29 AM.
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  9. #9
    MHF Contributor Unknown008's Avatar
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    Maybe taking another approach to have less of those square roots...

    ok, I'll try this.

    \int^{\frac{\pi}{3}}_0 (sin(x) - cos(x))^2 dx = \int^{\frac{\pi}{3}}_0 sin^2(x) - 2sin(x)cos(x) +  cos^2(x) dx

    = \int^{\frac{\pi}{3}}_0 1 - sin(2x)dx

    = [x+ \frac12 cos(2x)]^{\frac{\pi}{3}}_0

    = [\frac{\pi}{3} + \frac12 cos(\frac{2\pi}{3})] - [ 0 + \frac12 cos(0)]

    = \frac{\pi}{3} -\frac14 - \frac12

    = \frac{\pi}{3} -\frac34

    That's better
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  10. #10
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    i see my mistake, i divided by the integral!!!!
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  11. #11
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    Quote Originally Posted by Unknown008 View Post
    \int^{\frac{\pi}{3}}_0(sinx-cosx)^2 dx = [\frac{(sinx - cosx)^3}{3(cosx + sinx)}]^{\frac{\pi}{3}} _0

    =  \frac{(sin(\frac{\pi}{3}) - cos(\frac{\pi}{3}))^3}{3(cos(\frac{\pi}{3}) + sin(\frac{\pi}{3}))} -  \frac{(sin(0) - cos(0))^3}{3(cos(0) + sin(0))}

    =  \frac{(\frac{\sqrt{3}}{2} - \frac{1}{2})^3}{3(\frac12 + \frac{\sqrt{3}}{2})} -  \frac{(0 - 1)^3}{3(1 + 0)}

    =  \frac{(\frac{\sqrt{3}-1}{2})^3}{3(\frac{1+\sqrt{3}}{2})} -  \frac{-1}{3}
    this method probably cannot be used as the answer is wrong!!

    yr 2nd method is right

    any experts can clarify if it is wrong?
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  12. #12
    MHF Contributor Unknown008's Avatar
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    http://www.wolframalpha.com/input/?i=\int^{pi%2F3}_{0}+(sin(x)+-+cos(x))^2

    I just checked. The latter solution is supposed to be the correct one.

    So, I guess the first method is not the correct one
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