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

[Pi-Bond]

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.

 

[pmsrw3]

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.

 

[Pi-Bond]

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?

 

[pmsrw3]

Nope!

 

[CellCoree]

Yes, there is a way to evaluate the integral \int_0^T sin(\theta)dt solely from the information given. We can use the fact that the derivative of cos(\theta) is -sin(\theta) and apply the fundamental theorem of calculus, which states that the integral of a function's derivative over an interval is equal to the difference of the function evaluated at the endpoints of the interval. In this case, we can use the given information to rewrite the integral as:

\int_0^T sin(\theta)dt = -\int_0^T \frac{d}{dt}cos(\theta)dt = -[cos(\theta(T)) - cos(\theta(0))]

Since we know the value of the integral \int_0^T cos(\theta)dt and the function \theta(t), we can evaluate the endpoints of the interval and substitute them into the expression above. This will give us the value of the integral \int_0^T sin(\theta)dt. Therefore, with the given information, we can evaluate the integral \int_0^T sin(\theta)dt solely from the information given.