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Hr.Tboy
integration of x.cosx.sinx.dx for this period: -pi<x<pi ?
it should be solved by part integration! but i dont know which is dV and which is U?
\[\int xsin(x)cos(x)\] \(sin(x)cos(x) = 1/2sin(2x)\) So, \[\int x* \frac{1}{2}sin(2x) => \frac{1}{2} \int xsin(2x)\] You should be able to do it now.
sin(x)cos(x)=1/2sin(2x) how?
Integration by parts. \( u =x\) \(dv = sin(2x)\) Well, \(2sinxcosx = sin(2x)\) so \(sinxcosx = 1/2sin(2x)\)
@Mimi_x3 really good teacher;)
@Mimi_x3 so sinx is an oven function. and period is symmetric. so the answer is zero for 1part of part function. yea?
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.
sry. i mean odd and even function. sinx is an odd function!
http://www.wolframalpha.com/input/?i=integrate+xsinxcosx+from+-pi+to+pi
if you want to solve it visually..