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Math Help - intergration

  1. #1
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    intergration

    y= (x/ √ a^2 - x^2 ) - sin^-1 (x-a) (integrate)

    ∫ ln x /x^2 dx

    ∫ cos^6 1/2x dx

    umm how do i do these ! pleases show me with working

    thank you



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  2. #2
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    Re: intergration

    In the second expression, integrate by parts \displaystyle \int v u' = vu - \int uv'

    Use \displaystyle v = \ln(x) , u' = \frac{1}{x^2}
    Last edited by pickslides; September 26th 2012 at 06:41 PM.
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  3. #3
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    Re: intergration

    Hello, arsenal12345!

    Here's the third one . . .


    \int \cos^6\!\tfrac{x}{2}\,dx

    \cos^6\!\tfrac{x}{2} \;=\;\left(\cos^2\!\tfrac{x}{2}\right)^3 \;=\;\left(\tfrac{1 + \cos x}{2}\right)^3 \;=\;\tfrac{1}{8}\left(1 + 3\cos x + 3\cos^2\!x + \cos^3\!x\right)

    . . . . =\;\tfrac{1}{8}\left(1 + 3\cos x + 3\left[\tfrac{1+\cos2x}{2}\right] + \cos^2\!x\!\cdot\cos x\right)

    . . . . =\;\tfrac{1}{8}\left(1 + 3\cos x + \tfrac{3}{2} + \tfrac{3}{2}\cos2x + [1-\sin^2\!x]\cos x\right)

    . . . . =\;\tfrac{1}{8}\left(\tfrac{5}{2} + 3\cos x + \tfrac{3}{2}\cos2x + \cos x - \sin^2\!x\cos x\right)

    . . . . =\;\tfrac{1}{8}\left(\tfrac{5}{2} + 4\cos x +\tfrac{3}{2}\cos2x - \sin^2\!x\cos x\right)


    You should be able to handle those four integrals.
    Thanks from arsenal12345
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