`int[sinalphasin(x-alpha)+sin^2(x/2-alpha)]dx`

Answer

To solve the integral \( I = \int \left( \sin \alpha \sin(x - \alpha) + \sin^2\left(\frac{x}{2} - \alpha\right) \right) dx \), we can break it down into manageable parts. Here’s the step-by-step solution:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \left( \sin \alpha \sin(x - \alpha) + \sin^2\left(\frac{x}{2} - \alpha\right) \right) dx \]

### Step 2: Separate the Integral
We can separate the integral into two parts:
\[ I = \int \sin \alpha \sin(x - \alpha) \, dx + \int \sin^2\left(\frac{x}{2} - \alpha\right) \, dx \]

### Step 3: Integrate the First Part
For the first part, we can factor out \(\sin \alpha\) since it is a constant:
\[ I_1 = \sin \alpha \int \sin(x - \alpha) \, dx \]
The integral of \(\sin(x - \alpha)\) is:
\[ \int \sin(x - \alpha) \, dx = -\cos(x - \alpha) \]
Thus,
\[ I_1 = -\sin \alpha \cos(x - \alpha) \]

### Step 4: Integrate the Second Part
For the second part, we use the identity for \(\sin^2\):
\[ \sin^2 a = \frac{1 - \cos(2a)}{2} \]
Applying this to \(\sin^2\left(\frac{x}{2} - \alpha\right)\):
\[ \sin^2\left(\frac{x}{2} - \alpha\right) = \frac{1 - \cos\left(x - 2\alpha\right)}{2} \]
Thus, we can rewrite the integral:
\[ I_2 = \int \sin^2\left(\frac{x}{2} - \alpha\right) \, dx = \int \frac{1 - \cos\left(x - 2\alpha\right)}{2} \, dx \]
This can be separated as:
\[ I_2 = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos\left(x - 2\alpha\right) \, dx \]
The integrals evaluate to:
\[ \int 1 \, dx = x \quad \text{and} \quad \int \cos\left(x - 2\alpha\right) \, dx = \sin\left(x - 2\alpha\right) \]
Thus,
\[ I_2 = \frac{x}{2} - \frac{1}{2} \sin\left(x - 2\alpha\right) \]

### Step 5: Combine the Results
Now we combine \(I_1\) and \(I_2\):
\[ I = I_1 + I_2 = -\sin \alpha \cos(x - \alpha) + \frac{x}{2} - \frac{1}{2} \sin\left(x - 2\alpha\right) \]

### Step 6: Add the Constant of Integration
Finally, we add the constant of integration \(C\):
\[ I = -\sin \alpha \cos(x - \alpha) + \frac{x}{2} - \frac{1}{2} \sin\left(x - 2\alpha\right) + C \]

### Final Answer
Thus, the integral evaluates to:
\[ \int \left( \sin \alpha \sin(x - \alpha) + \sin^2\left(\frac{x}{2} - \alpha\right) \right) dx = -\sin \alpha \cos(x - \alpha) + \frac{x}{2} - \frac{1}{2} \sin\left(x - 2\alpha\right) + C \]