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Math Help - Ηelp for definite integral

  1. #1
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    Ηelp for definite integral

    Ηelp for definite integral
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Ηelp for definite integral

    V^2=\frac{1}{2\pi}\left \{\int_{0}^{\pi} \eta \mu^2 (vt)d(vt)\right \}^{\frac{1}{2}}

    Calculating the integral gives us:
    \int_{0}^{\pi} \eta \mu^2 (vt)d(vt)\right = \eta \mu^2 \int_{0}^{\pi} (vt)d(vt) = \eta \mu^2\left[\frac{(vt)^2}{2}\right]_{0}^{\pi} = \eta \mu^2\left(\frac{\pi^2}{2}\right)

    The answer should be clear now.
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    Re: Ηelp for definite integral

    Siron,

    I' m sure that what you wrote me does not hold. It is not allowed to seperate sin^2 from (ut), because sinf(x) is a function.

    I appreciate your taking the time to write to me but it would be better not to help, if you do not know.

    M.
    Last edited by mbempeni; May 29th 2012 at 10:58 AM.
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    Re: Ηelp for definite integral

    Perhaps it would help if you told us what in the world you are talking about! There is no "sin" or "sinf" in the attached file.
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    Re: Ηelp for definite integral

    You are right. Sorry.
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  6. #6
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    Re: Ηelp for definite integral

    power reduction identity ...

    \sin^2{u} = \frac{1-\cos(2u)}{2}

    ... so do it.
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    Re: Ηelp for definite integral

    Is it possible for you to write me the hole solution because I have a doubt for one point of the procedure?

    Thanks in advance.
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  8. #8
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    Re: Ηelp for definite integral

    Quote Originally Posted by mbempeni View Post
    Is it possible for you to write me the hole solution because I have a doubt for one point of the procedure?

    Thanks in advance.
    Write down the procedure you attempted, then someone may comment on whether the procedure you followed is correct ... or not.
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  9. #9
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    Re: Ηelp for definite integral

    I' ve just done it!
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  10. #10
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    Re: Ηelp for definite integral

    u = \omega t

    \int_0^\pi \sin^2{u} \, du

    \frac{1}{2} \int_0^\pi 1 - \cos(2u) \, du

    \frac{1}{2} \left[u - \frac{\sin(2u)}{2} \right]_0^\pi

    \frac{1}{2} \left[\left(\pi - 0 \right)-\left(0-0\right)\right] = \frac{\pi}{2}
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