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

• Pi-Bond
In summary, knowing the value of an integral \int_0^T cos(\theta)dt = x, and with the additional information that \theta = \theta(t), there is no way to solely evaluate the integral \int_0^T sin(\theta)dt. The absolute value can be obtained, but without further information about the function \theta(t), the integral cannot be evaluated.
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.

Last edited:
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!

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.

## 1. What is the purpose of using $\theta(t)$ in the integral?

The use of $\theta(t)$ in the integral allows for the integration to be evaluated over a specific interval, rather than the entire domain. This is useful in cases where the function being integrated has different behaviors or properties over different intervals.

## 2. How do you compute the integral with $\theta(t)$?

To compute the integral with $\theta(t)$, you first need to identify the interval over which the integration will be evaluated. Then, you can use the appropriate integration techniques for the function inside the integral, taking into account the presence of $\theta(t)$ in the limits of integration.

## 3. Can $\theta(t)$ be used in any type of integral?

Yes, $\theta(t)$ can be used in any type of integral as long as it is used in the limits of integration. It is commonly used in definite integrals, but it can also be used in indefinite integrals when the function being integrated has different behaviors over different intervals.

## 4. What are some common functions that use $\theta(t)$ in the integral?

Functions that are commonly evaluated using $\theta(t)$ in the integral include trigonometric functions (such as sin, cos, and tan), exponential functions, and piecewise-defined functions. These functions often have different behaviors or properties over different intervals, making it necessary to use $\theta(t)$ in the integral.

## 5. Are there any limitations or considerations when using $\theta(t)$ in the integral?

One limitation to consider when using $\theta(t)$ in the integral is the smoothness of the function being integrated. If the function has discontinuities or undefined points within the interval, the integral may not be well-defined. It is also important to carefully choose the interval over which the integration will be evaluated, as this can greatly affect the result.

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