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

1. anonymous

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

2. Mimi_x3

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

3. anonymous

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

4. Mimi_x3

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

5. anonymous

@Mimi_x3 thnx:)

6. anonymous

@Mimi_x3 really good teacher;)

7. Mimi_x3

:)

8. anonymous

@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

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. anonymous

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

11. Mimi_x3
12. Mimi_x3

if you want to solve it visually..

13. anonymous

ok

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