Evalueta
`int_(0)^(T)E_(0)I_(0)sin omega t sin(omegat+phi)dt`

Answer

To evaluate the integral

\[ I = \int_{0}^{T} E_0 I_0 \sin(\omega t) \sin(\omega t + \phi) \, dt, \]

we can use the product-to-sum identities for sine functions. The identity states that:

\[ \sin A \sin B = \frac{1}{2} \left[ \cos(A - B) - \cos(A + B) \right]. \]

### Step 1: Apply the product-to-sum identity

Let \( A = \omega t \) and \( B = \omega t + \phi \). Then we have:

\[ I = \int_{0}^{T} E_0 I_0 \cdot \frac{1}{2} \left[ \cos(\omega t - (\omega t + \phi)) - \cos(\omega t + (\omega t + \phi)) \right] dt. \]

This simplifies to:

\[ I = \frac{E_0 I_0}{2} \int_{0}^{T} \left[ \cos(-\phi) - \cos(2\omega t + \phi) \right] dt. \]

### Step 2: Simplify the integral

Since \(\cos(-\phi) = \cos(\phi)\), we can rewrite the integral as:

\[ I = \frac{E_0 I_0}{2} \left[ \cos(\phi) \int_{0}^{T} dt - \int_{0}^{T} \cos(2\omega t + \phi) dt \right]. \]

### Step 3: Evaluate the first integral

The first integral is straightforward:

\[ \int_{0}^{T} dt = T. \]

### Step 4: Evaluate the second integral

For the second integral, we have:

\[ \int_{0}^{T} \cos(2\omega t + \phi) dt. \]

Using the substitution \( u = 2\omega t + \phi \), we find:

\[ du = 2\omega dt \quad \Rightarrow \quad dt = \frac{du}{2\omega}. \]

Changing the limits accordingly, when \( t = 0 \), \( u = \phi \) and when \( t = T \), \( u = 2\omega T + \phi \). Thus, we have:

\[ \int_{\phi}^{2\omega T + \phi} \cos(u) \frac{du}{2\omega} = \frac{1}{2\omega} [\sin(u)]_{\phi}^{2\omega T + \phi}. \]

This evaluates to:

\[ \frac{1}{2\omega} \left[ \sin(2\omega T + \phi) - \sin(\phi) \right]. \]

### Step 5: Substitute back into the integral

Now substituting back into our expression for \( I \):

\[ I = \frac{E_0 I_0}{2} \left[ \cos(\phi) T - \frac{1}{2\omega} \left( \sin(2\omega T + \phi) - \sin(\phi) \right) \right]. \]

### Step 6: Substitute \( T = \frac{2\pi}{\omega} \)

Now we substitute \( T = \frac{2\pi}{\omega} \):

\[ I = \frac{E_0 I_0}{2} \left[ \cos(\phi) \frac{2\pi}{\omega} - \frac{1}{2\omega} \left( \sin(2\cdot 2\pi + \phi) - \sin(\phi) \right) \right]. \]

Since \(\sin(2\cdot 2\pi + \phi) = \sin(\phi)\), the second term simplifies to zero:

\[ I = \frac{E_0 I_0}{2} \left[ \cos(\phi) \frac{2\pi}{\omega} \right]. \]

### Final Result

Thus, the final result for the integral is:

\[ I = \frac{E_0 I_0 \pi \cos(\phi)}{\omega}. \]