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Trignometric Integrals

  1. Jul 12, 2011 #1
    Suppose I know the value of an integral:

    [itex]\int_0^T cos(\theta)dt = x[/itex]

    Is there any way to evaluate the integral [itex]\int_0^T sin(\theta)dt[/itex] solely from this information?

    EDIT: [itex]\theta=\theta(t)[/itex], i.e. [itex]\theta[/itex] is a function of t.
    Last edited: Jul 12, 2011
  2. jcsd
  3. Jul 12, 2011 #2
    Well, you can get the absolute value.

    \int_0^T \cos(\theta) dt = T \cos(\theta) = x \\
    \cos(\theta) = x/T \\
    \cos^2(\theta) + \sin^2(\theta) = 1 \\
    sin(\theta) = \pm\sqrt{1-\cos^2(\theta)} \\
    \int_0^T \sin(\theta) dt = T \sin(\theta) \\

    Should be obvious from there.
  4. Jul 12, 2011 #3
    Unfortunately, I omitted a key piece of information in my original post - [itex]\theta[/itex] is a function of t, i.e. [itex]\theta=\theta(t)[/itex]. Is there any way then?
  5. Jul 12, 2011 #4
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