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

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Discussion Overview

The discussion revolves around evaluating the integral \(\int_0^T \sin(\theta) dt\) given the value of the integral \(\int_0^T \cos(\theta) dt = x\), where \(\theta\) is a function of \(t\). The scope includes mathematical reasoning and exploration of relationships between the integrals of sine and cosine functions when the angle is variable.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant suggests that knowing \(\int_0^T \cos(\theta) dt = x\) allows for the evaluation of \(\int_0^T \sin(\theta) dt\) by using trigonometric identities, assuming \(\theta\) is constant.
  • Another participant points out that since \(\theta\) is a function of \(t\), this assumption does not hold, implying that the relationship between the integrals cannot be established in the same way.
  • A later reply asserts that it is not possible to evaluate \(\int_0^T \sin(\theta) dt\) given the information provided, emphasizing the dependency on \(\theta(t)\).

Areas of Agreement / Disagreement

Participants express disagreement regarding the ability to evaluate the sine integral based on the cosine integral when \(\theta\) is a function of \(t\). There is no consensus on a method to derive \(\int_0^T \sin(\theta) dt\) from the given information.

Contextual Notes

The discussion highlights the importance of the functional form of \(\theta(t)\) and its implications for the integrals involved. The assumptions made in the initial responses do not hold under the condition that \(\theta\) varies with \(t\.

Pi-Bond
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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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Well, you can get the absolute value.

[tex] \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*}[/tex]

Should be obvious from there.
 
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?
 
Nope!
 

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