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anonymous
 3 years ago
integration of x.cosx.sinx.dx for this period: pi<x<pi ?
anonymous
 3 years ago
integration of x.cosx.sinx.dx for this period: pi<x<pi ?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it should be solved by part integration! but i dont know which is dV and which is U?

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.5\[\int xsin(x)cos(x)\] \(sin(x)cos(x) = 1/2sin(2x)\) So, \[\int x* \frac{1}{2}sin(2x) => \frac{1}{2} \int xsin(2x)\] You should be able to do it now.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sin(x)cos(x)=1/2sin(2x) how?

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.5Integration by parts. \( u =x\) \(dv = sin(2x)\) Well, \(2sinxcosx = sin(2x)\) so \(sinxcosx = 1/2sin(2x)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Mimi_x3 really good teacher;)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Mimi_x3 so sinx is an oven function. and period is symmetric. so the answer is zero for 1part of part function. yea?

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.5What 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sry. i mean odd and even function. sinx is an odd function!

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.5http://www.wolframalpha.com/input/?i=integrate+xsinxcosx+from+pi+to+pi

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.5if you want to solve it visually..
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