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Math Help - Integration By Parts

  1. #1
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    Integration By Parts

    Could someone please do me a favor and show me the steps to integrate this problem: 2PI fnInt((x)(-10cos((PI/6.44)x)+18 - 1 ),x,0,0.2)? Thanks.
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  2. #2
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    Re: Integration By Parts

    Hey knag95.

    It is usually a better idea for the person attempting the question to show what they have tried rather than to just be handed the answer. Can you show us what you tried?
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  3. #3
    Senior Member x3bnm's Avatar
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    Re: Integration By Parts

    We can simplify to write:

    \begin{align*}2*\pi \int_{0}^{0.2} x * \left(-10 \cos \left(\frac{\pi * x}{6.44}\right) + 18 -1\right) \,\,dx = 2 * \pi \int_{0}^{0.2} x * & \Big(-10 \cos \left(\frac{\pi * x}{6.44}\right)\\ &+ 17\Big)\,\,dx\end{align*}

    Now we know that if u(x) and v(x) are two functions then:

    \int \frac{d(u(x)v(x))}{dx} dx = u(x)v(x) = \int u(x) dv + \int v(x) du

    If we rearrange this we get:

    \int u(x) dv = u(x)v(x) - \int v(x) du.......\text{Equation(1)}

    Let u(x) = x \text{ then } \frac{du}{dx} = 1 \;\;\therefore du = dx and let:

    dv = \left(-10 \cos \left(\frac{\pi * x}{6.44}\right) + 17\right)\,\,dx

    then by integrating both sides:


    v(x) = \frac{-10 * 6.44 \sin \left(\frac{\pi * x}{6.44}\right)}{\pi}+ 17x



    Now by plugging in appropriate values to Equation(1) we get:

    \begin{align*}& 2*\pi \int_{0}^{0.2} x * (-10 \cos (\frac{\pi * x}{6.44}) + 17) \,\,dx\\ =& 2*\pi\Big[x * \left(\frac{-10 * 6.44 \sin (\frac{\pi * x}{6.44})}{\pi}+ 17x \right) - \int_{0}^{0.2} \left(\frac{-10 * 6.44 \sin (\frac{\pi * x}{6.44})}{\pi}+ 17x\right)\,\,dx\Big]\\=& 2*\pi \Big[ x * \left(\frac{-10 * 6.44 \sin (\frac{\pi * x}{6.44})}{\pi}+ 17x \right) -  \left(\frac{10 * (6.44)^2 \cos (\frac{\pi * x}{6.44})}{(\pi)^2}+ \frac{17x^2}{2} \right)\Big]_{0}^{0.2}\Big]\\=& 2* \pi\Big[0.2 * \left(\frac{-10 * 6.44 \sin (\frac{\pi * 0.2}{6.44})}{\pi}+ (17*0.2) \right) - \left(\frac{10 * (6.44)^2 \cos (\frac{\pi * 0.2}{6.44})}{(\pi)^2}+ \frac{17*0.2^2}{2}  \right)\\&  - 0 + \left(\frac{10 * (6.44)^2}{(\pi)^2}+ 0\right)\Big]\\=& 2*\pi*(0.280634-42.161701+42.021542)\\=& 0.88263 \end{align*}

    That's the answer. Hope it helps.

    You can check the answer at: integrate (2*pi * ((x)(-10 * cos((pi/6.44) * x) +17 ))) dx from x=0 to 0.2 - Wolfram|Alpha
    Last edited by x3bnm; March 30th 2013 at 04:18 PM.
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