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

Question

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

anonymous · Answer

it should be solved by part integration! but i dont know which is dV and which is U?

mimi_x3 · Answer

$$\int xsin(x)cos(x)$$
$sin(x)cos(x) = 1/2sin(2x)$
So, 
$$\int x* \frac{1}{2}sin(2x) => \frac{1}{2} \int xsin(2x)$$
You should be able to do it now.

anonymous · Answer

sin(x)cos(x)=1/2sin(2x) how?

mimi_x3 · Answer

Integration by parts.
$ u =x$ 
$dv = sin(2x)$

Well, $2sinxcosx = sin(2x)$ so $sinxcosx = 1/2sin(2x)$

anonymous · Answer

@Mimi_x3  thnx:)

anonymous · Answer

@Mimi_x3  really good teacher;)

mimi_x3 · Answer

:)

anonymous · Answer

@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 · Answer

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.

anonymous · Answer

sry. i mean odd and even function. sinx is an odd function!

mimi_x3 · Answer

http://www.wolframalpha.com/input/?i=integrate+xsinxcosx+from+-pi+to+pi

mimi_x3 · Answer

if you want to solve it visually..

anonymous · Answer

ok