# Trignometric Integrals

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.

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

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!