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Member warned about posting without the homework template

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- Thread starter core1985
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- #1

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Member warned about posting without the homework template

- #2

BvU

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The template is there for a reason, don't erase it but use it; it will be to your benefit.

What is the question ? and what is the relationship between your first line and the second ?

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- #4

BvU

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But the second line does not reflect that ( it says ##\ \sin (2kx) \ ## instead of ##\ \sin^2 (kx) \ ## ).

So :

What is the question ? and what is the relationship between your first line and the second ?

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yes yes it is sin^2(kx) we can use 1-cos2(x)/2 formula here

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but that nasty exponential how to handle that

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it is liboff problem 3.15 I have found <p> that is zero but I stuck at A???

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how to integrate E^x^2/a^2 sinkx

- #11

Mark44

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For the latter expression, if you mean ##\frac{1 - \cos^2(x)}2##, use parentheses around the terms in the numerator. What you wrote means ##1 - \frac{\cos(2x)}2##. In any case, ##\sin^2(kx) \ne \frac{1 - \cos(2x)}{2}##. You have to consider that k mulitplier.yes yes it is sin^2(kx) we can use 1-cos2(x)/2 formula here

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thanks I am new to this website

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BvU

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Otherwise there is CRC handbook of chemistry and physics, or Abramowitz (7.4.6)

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one thing more can I change sin(kx) in to exponentionals and then try to solve will it work or not??

- #16

BvU

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IF the function is even (##\ f(x) = f(-x)\ ##) then yes.

You can give it a try...one thing more can I change sin(kx) in to exponentionals and then try to solve will it work or not??

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what do you suggest now changing sin to exponential using euler formula or use this

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but cos(kx) is even ?? so I can use this to solve this nasty integral

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ok I am solving it by both methods and will tell you what I got

- #20

BvU

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Yes you canbut cos(kx) is even ?? so I can use this to solve this nasty integral

That would be the idea. But it doesn't look clean and quick to me, such a complex exponential...what do you suggest now changing sin to exponential using euler formula or use this

After all, integrating ##\ e^{-x^2}\ ## alone already requires ingenious mathematical manipulating...

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- #22

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