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Math Help - Integration

  1. #1
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    Integration

    It is given that . Find the values of the constants and .

    How would you solve this?
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  2. #2
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    Let I = \int_0^{\frac{\pi}{2}} e^{4y} \sin (3y) dy

    Use integration by parts twice.

    For the first round, use: u = e^{4y} and dv =\sin (3y) dy

    To get: I  = -\frac{1}{3}e^{4y}\cos (3y) \bigg|_0^{\frac{\pi}{2}} + \frac{4}{3} {\color{blue}\int_0^{\frac{\pi}{2}} e^{4y} \cos (3y) dy}

    Use integrationg by parts again on the blue using: u = e^{4y} and dv = \cos (3y) dy

    To get: I = -\frac{1}{3}e^{4y}\cos (3y)\bigg|_0^{\frac{\pi}{2}} + \frac{4}{3} \left(\frac{1}{3} e^{4y} \sin (3y) \bigg|_0^{\frac{\pi}{2}} - \frac{4}{3} \int_0^{\frac{\pi}{2}} e^{4y} \sin (3y) dy \right)

    You should notice that you get I on the right hand side again. So juse use a little bit of algebra to solve for I:

    {\color{red}I} = -\frac{1}{3}e^{4y}\cos (3y)\bigg|_0^{\frac{\pi}{2}} + \frac{4}{3} \left(\frac{1}{3} e^{4y} \sin (3y) \bigg|_0^{\frac{\pi}{2}} - \frac{4}{3} {\color{red}I} \right)
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  3. #3
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    Hello, Haris!

    It is given that: . \int^{\frac{\pi}{2}}_0 e^{4x}\sin(3x)\,dx \;=\; pe^{2\pi} + q

    Find the values of the constants p and q.

    Obviously, we do the integration first . . .

    . . \int^{\frac{\pi}{2}}_0e^{4x}\sin(3x)\,dx\;=\;\frac  {e^{4x}}{25}\bigg[4\sin(3x) - 3\cos(3x)\bigg]^{\frac{\pi}{2}}_0 ..
    This takes more work than I want to show here


    Evaluate: . \frac{e^{4(\frac{\pi}{2})}}{25}   \bigg[4\sin\tfrac{3\pi}{2} -3\cos\tfrac{3\pi}{2}\bigg] - \frac{e^0}{25}\bigg[4\sin0 - 3\cos0 \bigg]

    . . . . . . = \;\frac{e^{2\pi}}{25}\bigg[4(-1) - 3(0)\bigg] - \frac{1}{25}\bigg[4(0) - 3(1)\bigg]

    . . . . . . = \;\frac{e^{2\pi}}{25}(-4) - \frac{1}{25}(-3)

    . . . . . . =\;-\frac{4}{25}e^{2\pi} + \frac{3}{25}


    Therefore: . p \:=\:-\frac{4}{25},\;\;q \:=\:\frac{3}{25}

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  4. #4
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    \int^\frac{\pi}{2}_0 e^{4x}sin(3x) dx

    but sin(3x)=Im[e^{3ix}] so,

    \int^\frac{\pi}{2}_0 e^{4x}e^{3ix} dx

    \int^\frac{\pi}{2}_0 e^{(4+3i)x} dx

    I=\bigg[\frac{1}{4+3i} e^{(4+3i)x}\bigg]^\frac{\pi}{2}_0

    I=\bigg[\frac{1}{25} (4-3i)e^{(4+3i)x}\bigg]^\frac{\pi}{2}_0

    I=\bigg[\frac{1}{25} (4-3i)e^{4x}e^{3ix}\bigg]^\frac{\pi}{2}_0

    I=\bigg[\frac{1}{25}e^{4x} (4-3i)(cos(3x)+isin(3x))\bigg]^\frac{\pi}{2}_0

    I=\bigg[\frac{1}{25}e^{4x}(4cos(3x)+4isin(3x)-3icos(3x)+3sin(3x))\bigg]^\frac{\pi}{2}_0

    Now take the imaginary part of the result:

    I=\bigg[\frac{1}{25}e^{4x}(4sin(3x)-3cos(3x))\bigg]^\frac{\pi}{2}_0

    Is this method correct?
    Last edited by Haris; January 28th 2009 at 07:00 AM.
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  5. #5
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    Yes, that's another way of doing it.

    When (in particular) studying the improper integral \int_0^\infty e^{-\alpha x}\sin(\beta x)\,dx its calculation is very easy by using that method.
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