Evalueta
`int_(0)^(T)I_(0)sin omegat dt`

Answer

To evaluate the integral

\[ \int_{0}^{T} I_{0} \sin(\omega t) \, dt, \]

we can follow these steps:

### Step 1: Factor out the constant
Since \(I_{0}\) is a constant, we can factor it out of the integral:

\[ I_{0} \int_{0}^{T} \sin(\omega t) \, dt. \]

### Step 2: Integrate \(\sin(\omega t)\)
The integral of \(\sin(\omega t)\) with respect to \(t\) is given by:

\[ \int \sin(\omega t) \, dt = -\frac{1}{\omega} \cos(\omega t) + C, \]

where \(C\) is the constant of integration. Therefore, we can write:

\[ \int_{0}^{T} \sin(\omega t) \, dt = \left[-\frac{1}{\omega} \cos(\omega t)\right]_{0}^{T}. \]

### Step 3: Evaluate the definite integral
Now we need to evaluate this expression from \(0\) to \(T\):

\[ -\frac{1}{\omega} \cos(\omega T) - \left(-\frac{1}{\omega} \cos(0)\right). \]

Since \(\cos(0) = 1\), we have:

\[ -\frac{1}{\omega} \cos(\omega T) + \frac{1}{\omega}. \]

### Step 4: Combine the results
Now we can combine the results:

\[ \frac{1}{\omega} - \frac{1}{\omega} \cos(\omega T) = \frac{1}{\omega} (1 - \cos(\omega T)). \]

### Step 5: Multiply by \(I_{0}\)
Finally, we multiply by \(I_{0}\):

\[ I_{0} \left(\frac{1}{\omega} (1 - \cos(\omega T})\right). \]

### Final Result
Thus, the evaluated integral is:

\[ \int_{0}^{T} I_{0} \sin(\omega t) \, dt = \frac{I_{0}}{\omega} (1 - \cos(\omega T)). \]