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Hr.Tboy

  • 3 years ago

integration of x.cosx.sinx.dx for this period: -pi<x<pi ?

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  1. Hr.Tboy
    • 3 years ago
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    it should be solved by part integration! but i dont know which is dV and which is U?

  2. Mimi_x3
    • 3 years ago
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    \[\int xsin(x)cos(x)\] \(sin(x)cos(x) = 1/2sin(2x)\) So, \[\int x* \frac{1}{2}sin(2x) => \frac{1}{2} \int xsin(2x)\] You should be able to do it now.

  3. Hr.Tboy
    • 3 years ago
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    sin(x)cos(x)=1/2sin(2x) how?

  4. Mimi_x3
    • 3 years ago
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    Integration by parts. \( u =x\) \(dv = sin(2x)\) Well, \(2sinxcosx = sin(2x)\) so \(sinxcosx = 1/2sin(2x)\)

  5. Hr.Tboy
    • 3 years ago
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    @Mimi_x3 thnx:)

  6. Hr.Tboy
    • 3 years ago
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    @Mimi_x3 really good teacher;)

  7. Mimi_x3
    • 3 years ago
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    :)

  8. Hr.Tboy
    • 3 years ago
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    @Mimi_x3 so sinx is an oven function. and period is symmetric. so the answer is zero for 1part of part function. yea?

  9. Mimi_x3
    • 3 years ago
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    What is an "oven" function? If you solving through the intervals; im not sure.. Why not integrate it; and sub in the limits and to see the area.

  10. Hr.Tboy
    • 3 years ago
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    sry. i mean odd and even function. sinx is an odd function!

  11. Mimi_x3
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=integrate+xsinxcosx+from+-pi+to+pi

  12. Mimi_x3
    • 3 years ago
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    if you want to solve it visually..

  13. Hr.Tboy
    • 3 years ago
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    ok

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