Trignometric Integrals

  • Thread starter Pi-Bond
  • Start date
  • #1
302
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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.
 
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Answers and Replies

  • #2
459
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Well, you can get the absolute value.

[tex]
\begin{eqnarray*}
\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) \\
\end{eqnarray*}
[/tex]

Should be obvious from there.
 
  • #3
302
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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?
 
  • #4
459
0
Nope!
 

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