core1985
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BvU said:What is the question ? and what is the relationship between your first line and the second ?
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.core1985 said:yes yes it is sin^2(kx) we can use 1-cos2(x)/2 formula here
IF the function is even (##\ f(x) = f(-x)\ ##) then yes.core1985 said:so if I use number 6 then can I change limits to 0 to infinity multiplied by 2 then It can be applied ?
You can give it a try...core1985 said:one thing more can I change sin(kx) into exponentionals and then try to solve will it work or not??
Yes you cancore1985 said:but 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...core1985 said:what do you suggest now changing sin to exponential using euler formula or use this