anonymous
  • anonymous
Can someone explain this to me:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1377725540104:dw|
Luigi0210
  • Luigi0210
Uh, what is this? >.>
anonymous
  • anonymous
linear algebra.. orthogonality..

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blockcolder
  • blockcolder
Which part didn't you understand?
anonymous
  • anonymous
how did it become 0.I know that they are supposed to be "perpendicular"
anonymous
  • anonymous
everything I guess
anonymous
  • anonymous
it is zero because that is what you get when you integrate
blockcolder
  • blockcolder
It's probably better to use 'orthogonal' rather than 'perpendicular', when we're not talking about geometry.
anonymous
  • anonymous
the anti derivative of \[\sin(x)\cos(x)\] is \(\frac{1}{2}\sin^2(x)\)
anonymous
  • anonymous
how did it become 1/2sin2t?
anonymous
  • anonymous
plug in \(\pi\) you get \(0\), plug in \(-\pi\) you also get \(0\)
anonymous
  • anonymous
equation is not showing.Please draw...
anonymous
  • anonymous
one way is to put \(u=\sin(x), du =\cos(x)dx\)
anonymous
  • anonymous
refresh your browser
anonymous
  • anonymous
or in english a u substitution put u as sine, du as cosine and you get it in one step
anonymous
  • anonymous
I did...it won't work
Loser66
  • Loser66
you know that sin (2x) = 2 sinx cos x ; Your integrand is sinx cos x = 1/2 sin (2x) , right? \[\frac{1}{2}\int_{-\pi}^{\pi}sin (2x) dx = \frac{1}{4} cos (2x)\text{ from -\(\pi\) to \(\pi\)}\] just plug the value of the limit in, you got 0
Loser66
  • Loser66
hehe, ignore the last sentence, just do as usual. I am sorry for my carelessly stated it .
anonymous
  • anonymous
the equation is not showing

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