Thanks for that, but from here I am struggling to find the right identity to simplify equation!
This is not my problem but I can't help be intrigued on how to do it!
You're finding a0 for sin 6t, not sin |6t|, which is what the problem asked for.
For sin |6t| we ignore the - values and we are left with:
1/2pi [-1/6] = -1/12pi
Hope that is right!
thanks for that vela, i looked at the identities prior to you post but I didnt know how to apply them to this case. Would you have any pointers for which one it is? :)
Usually for integration of something like sinxcosx i use the identity:
sin2x = 2sinxcosx
But not sure how that applies in this case. I am very curious, hopefully someone can answer!
For an just substituting in I get:
(1/L) int [ sin(6t) * cos(nt) ] .dt
Not sure if you started with this...
If you did I am not sure how this led to:
the sin(nt)cos(6t) integral.
Interesting problem hmm, can you show how you got a0 and explain hoe you got upto where you are?
From there I think some trigonometric identities will be useful. But I will put pen to paper once I see where and how you got there!
Seems like a tricky question!
Good question, don't think I've done anything wrong to that point.
Basically I am trying to evaluate the following double integral, by changing the order of integration first.
cosh(x^2).dx.dy for 3y < or equal to: x < or equal to 3
and 0 < or equal to: y < or equal to 1
I changed the...
Homework Statement
Just trying to figure out the anti-derivative of cosh(x^2).
Homework Equations
I knowthe antiderivative cannot be expressed as an elementary function but I am pretty clueless of getting the antiderivative though!
The Attempt at a Solution
I am baffled by this one...
Sorry didn't even pick up on that, just a careless error! So it should be:
(e^-t)*[(L^-1)[3/(s^2 - 2^2)-(L^-1)[4s/(s^2 - 2^2)]]
(e^-t)*([3/2](L^-1)[2/(s^2 - 2^2)*-4(L^-1)[s/(s^2 - 2^2)])
From here I used the two following Laplace transformations:
a/(s^2 + a^2) = sin(at) and s/(s^2 + a^2)...
Thanks, I didn't see that at all!
From there I have tried to now split the expression into 2 parts to use the Laplace Transformation table, by doing:
(e^-t)*(L^-1)[3/(s^2 - 2^2)*(L^-1)[-4s/(s^2 - 2^2)]
(e^-t)*[3/2](L^-1)[2/(s^2 - 2^2)*-4(L^-1)[s/(s^2 - 2^2)]
From here I used the two...
I've simplified the numerator to get:
(7-4s) / [(s-1)^2 - 4]
From here I suspect I should break the expression up into 2 parts and compare them to the table of Laplace transformations (from other questions I've done). But I just can't find the right parts and Laplace transformations.
Thanks a lot for the replies guys, really appreciate it. Hmm I just made a small mistake in isolating L[y], I got that bit now thanks to your help!
I have worked at it and did get:
[3-4(s-1)] / [(s-1)^2 - 4]
It's what I do from here to get y(t) that's confusing me now. I can't seem to find a...