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Math Help - simple integral

  1. #1
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    simple integral

    trying ti do a problem

    I need the intregal of [(3/2)-2cos(2x)+((1/2)cos(4x)]

    i got (3/2)x -4sin(2x) + 2cos(4x)+c

    is that right?
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  2. #2
    Senior Member eumyang's Avatar
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    I'm getting
    \dfrac{3}{2}x - \sin\, 2x + \dfrac{1}{8}\sin\, 4x + C

    When you do u substitution on the second term, for instance:
    \begin{aligned}<br />
u &= 2x \\<br />
du &= 2 dx \\<br />
\int 2 \cos\,2x\; dx &= \int \cos\,u\;du \\<br />
&= \sin\,u + C \\<br />
&= \sin\,2x + C<br />
\end{aligned}
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  3. #3
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    thanks,

    so for the rest of the problem, I take it from the bonds 0 to pi.

    I got pi/4 [(3/2)pi-sin92pi)+(1/8)sin(4pi)]- (pi/4)[0-sin(0)-(1/8)sin(0)]

    what is the final answer to this?
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  4. #4
    Senior Member eumyang's Avatar
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    I'm not sure what you're doing, but if this is a definite integral going from 0 to pi, then you would have
    \left( \dfrac{3}{2}(\pi) - \sin\, 2(\pi) + \dfrac{1}{8}\sin\, 4(\pi) \right) - \left( \dfrac{3}{2}(0) - \sin\, 2(0) + \dfrac{1}{8}\sin\, 4(0) \right)
    = \dfrac{3\pi}{2}
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  5. #5
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    yeah urs looks like what I have, except there is a pi/4 outside the integral. how do I put that in? I have to put multiply it to both ()'s before subtracting them right?
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  6. #6
    Senior Member eumyang's Avatar
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    Grr, please state the ENTIRE problem next time. So I'm guessing that the problem was this:
    \dfrac{\pi}{4} {\displaystyle \int_0^{\pi}} \dfrac{3}{2} - 2\cos\, 2x + \dfrac{1}{2}\cos\, 4x\; dx

    After taking the integral you get:
    \dfrac{\pi}{4} \left[ \dfrac{3}{2}x - \sin\, 2x + \dfrac{1}{8}\sin\, 4x\; \Big{\vert}_0^{\pi} \right]

    Then evaluate:
    \dfrac{\pi}{4} \left[ \left( \dfrac{3}{2}(\pi) - \sin\, 2(\pi) + \dfrac{1}{8}\sin\, 4(\pi)  \right) - \left( \dfrac{3}{2}(0) - \sin\, 2(0) + \dfrac{1}{8}\sin\, 4(0)  \right) \right]
    = \dfrac{\pi}{4} \left( \dfrac{3\pi}{2} \right)
    = \dfrac{3\pi^2}{8}
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