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Math Help - Solve Integral Analytically

  1. #1
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    Solve Integral Analytically

    Integral from -2 to 3 of e^(2x)cos(3x)dx

    I think I am suppose to solve for the unknown integral using Integration by parts formula however I have tried multiplies times but no luck lol.
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Cohsty243 View Post
    Integral from -2 to 3 of e^(2x)cos(3x)dx

    I think I am suppose to solve for the unknown integral using Integration by parts formula however I have tried multiplies times but no luck lol.
    I = \int {{e^{2x}}\cos \left( {3x} \right)dx}  = \frac{{{e^{2x}}}}{3}\sin \left( {3x} \right) - \frac{2}{3}\int {{e^{2x}}\sin \left( {3x} \right)dx}  =

    = \frac{{{e^{2x}}}}{3}\sin \left( {3x} \right) + \frac{{2{e^{2x}}}}{9}\cos \left( {3x} \right) - \frac{4}{9}\int {{e^{2x}}\cos \left( {3x} \right)dx} .

    I = \frac{{{e^{2x}}}}{3}\sin \left( {3x} \right) + \frac{{2{e^{2x}}}}{9}\cos \left( {3x} \right) - \frac{4}{9}I \Leftrightarrow

    \Leftrightarrow I + \frac{4}{9}I = \frac{{{e^{2x}}}}{3}\sin \left( {3x} \right) + \frac{{2{e^{2x}}}}{9}\cos \left( {3x} \right) \Leftrightarrow

    \Leftrightarrow \frac{{13}}{9}I = \frac{{{e^{2x}}}}{9}\left[ {3\sin \left( {3x} \right) + 2\cos \left( {3x} \right)} \right] \Leftrightarrow

    \Leftrightarrow \boxed{I = \frac{{{e^{2x}}}}{{13}}\left[ {3\sin \left( {3x} \right) + 2\cos \left( {3x} \right)} \right] + C}

    Please note that the function f\left( x \right) = {e^{2x}}\cos \left( {3x} \right) has five roots

    {x_1} =  - \frac{\pi }{2};{\text{ }}{x_2} =  - \frac{\pi }{6};{\text{ }}{x_3} = \frac{\pi }{6};{\text{ }}{x_4} = \frac{\pi }{2};{\text{ }}{x_5} = \frac{{5\pi }}{6}

    on the interval - 2 \leqslant x \leqslant 3, ie, this function 6 times changes its sign in this interval. I hope you understand why this is necessary.

    If you calculate everything correctly, then your response should be approximately equal to 117.086.
    Last edited by DeMath; February 20th 2009 at 05:11 AM. Reason: typo
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    Thank you
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