Solving Integral Troubles with Vibrations Problem

  • Thread starter Saladsamurai
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In summary: But you cannot integrate dk/dy to get the value of k, as that would mean you are integrating a constant which would give you an infinite number of solutions. You need more information about the problem to solve for k.
  • #1
Saladsamurai
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Homework Statement



Alrighty. I have reduced a vibrations problem to an integral and I am having some trouble
evaluating it.

I have a value for t and need to find:[tex]-c\omega^2Z^2\int_0^t\cos^2(\omega t-\phi)\ dt[/tex] (1)I guess it is just my memory that is the problem.

If I had [itex]\int\cos^2(x)\ dx[/itex] It would not be a problem.

I am thinking now that I type this that a simple U substitution should do the trick right?EDIT:

If I let [itex]u=\omega t-\phi\ \Rightarrow du=\omega\ dt[/itex]

So (1) becomes:

[tex]-c\omega Z^2\int_0^t\cos^2u\ du[/tex]

Yes?
 
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  • #2
hi - i think your limits should become u(0) & u(t) as well
 
  • #3
If you just use the the double angle formula
[tex]\cos^2 t = \frac{1}{2} (1+\cos 2t) [/itex]
from the outset (because it's the next step after your u-substitution anyway), then you really don't need to do a u-substitution if [itex]\omega[/itex] and [itex]\phi[/itex] are just constants.
 
  • #4
pls help me too...integration problem

hi... i am new here and i hope someone can please answer my question too..

x-1 + dk/dy = x-1
so when we cancel both x-1 we get dk/dy = 0

my question is can i integrate dk/dy to get the k's value??

if i integrate dk/dy, am i getting C (constant) for the k value?

thank u very much...
 
  • #5
Yes, if dk/dy = 0, then k = a constant.
 

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