Evalueta
`int_(0)^(T//2)I_(0)sin omegatdt`
Given that `omegaT=2pi.`

Answer

To evaluate the integral

\[ \int_{0}^{\frac{T}{2}} I_0 \sin(\omega t) \, dt \]

given that \(\omega T = 2\pi\), we can follow these steps:

### Step 1: Identify the limits of integration and constants
The limits of integration are from \(0\) to \(\frac{T}{2}\). We know that \(\omega T = 2\pi\), which implies that

\[ T = \frac{2\pi}{\omega}. \]

Thus,

\[ \frac{T}{2} = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}. \]

### Step 2: Rewrite the integral
Substituting the limits into the integral, we have:

\[ \int_{0}^{\frac{\pi}{\omega}} I_0 \sin(\omega t) \, dt. \]

### Step 3: Factor out the constant
Since \(I_0\) is a constant, we can factor it out of the integral:

\[ I_0 \int_{0}^{\frac{\pi}{\omega}} \sin(\omega t) \, dt. \]

### Step 4: Perform the integration
To integrate \(\sin(\omega t)\), we use the formula:

\[ \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C. \]

Here, \(k = \omega\). Therefore, we have:

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

### Step 5: Evaluate the definite integral
Now we evaluate the definite integral from \(0\) to \(\frac{\pi}{\omega}\):

\[ I_0 \left[-\frac{1}{\omega} \cos(\omega t) \right]_{0}^{\frac{\pi}{\omega}}. \]

Calculating the limits:

1. At \(t = \frac{\pi}{\omega}\):

\[ -\frac{1}{\omega} \cos\left(\omega \cdot \frac{\pi}{\omega}\right) = -\frac{1}{\omega} \cos(\pi) = -\frac{1}{\omega} \cdot (-1) = \frac{1}{\omega}. \]

2. At \(t = 0\):

\[ -\frac{1}{\omega} \cos(0) = -\frac{1}{\omega} \cdot 1 = -\frac{1}{\omega}. \]

### Step 6: Combine the results
Now, substituting these values back into the integral:

\[ I_0 \left(\frac{1}{\omega} - \left(-\frac{1}{\omega}\right)\right) = I_0 \left(\frac{1}{\omega} + \frac{1}{\omega}\right) = I_0 \cdot \frac{2}{\omega}. \]

### Final Result
Thus, the value of the integral is:

\[ \int_{0}^{\frac{T}{2}} I_0 \sin(\omega t) \, dt = \frac{2 I_0}{\omega}. \]