Evaluating Integral with $\theta(t)$ - Sin | Cos

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The discussion centers on evaluating the integral of sin(θ(t)) given the integral of cos(θ(t)) over the same interval. Initially, it was suggested that the relationship between the two integrals could be derived using trigonometric identities. However, it was clarified that since θ is a function of t, the direct evaluation of the integral of sin(θ(t)) cannot be determined solely from the integral of cos(θ(t)). The conclusion is that without additional information about the function θ(t), it is not possible to evaluate the integral of sin(θ(t)). Thus, the relationship between the two integrals breaks down when θ is variable.
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Suppose I know the value of an integral:

\int_0^T cos(\theta)dt = x

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

EDIT: \theta=\theta(t), i.e. \theta is a function of t.
 
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Well, you can get the absolute value.

<br /> \begin{eqnarray*}<br /> \int_0^T \cos(\theta) dt = T \cos(\theta) = x \\<br /> \cos(\theta) = x/T \\<br /> \cos^2(\theta) + \sin^2(\theta) = 1 \\<br /> sin(\theta) = \pm\sqrt{1-\cos^2(\theta)} \\<br /> \int_0^T \sin(\theta) dt = T \sin(\theta) \\<br /> \end{eqnarray*}<br />

Should be obvious from there.
 
Unfortunately, I omitted a key piece of information in my original post - \theta is a function of t, i.e. \theta=\theta(t). Is there any way then?
 
Nope!
 
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