Recent content by WrittenStars

Yeah that didnt seem right. The problem I'm having is that very first step! LOL

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...

i am trying to solve a double integral question but I can't figure out how to anti-differentiate cosh(x^2). it's quite frustrating!

Sorry but isn't the chain rule for differentiating, not antidifferentiating?

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...