`int (2-x) sin x dx`=_____

Answer

To solve the integral \( \int (2 - x) \sin x \, dx \), we can break it down into two separate integrals. Here’s the step-by-step solution:

### Step 1: Break down the integral
We can express the integral as:
\[ I = \int (2 - x) \sin x \, dx = \int 2 \sin x \, dx - \int x \sin x \, dx \]

### Step 2: Solve the first integral
The first integral is straightforward:
\[ \int 2 \sin x \, dx = 2 \int \sin x \, dx = 2 (-\cos x) = -2 \cos x \]

### Step 3: Solve the second integral using integration by parts
For the second integral \( \int x \sin x \, dx \), we will use integration by parts. Let:
- \( u = x \) (which means \( du = dx \))
- \( dv = \sin x \, dx \) (which means \( v = -\cos x \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int x \sin x \, dx = -x \cos x - \int -\cos x \, dx \]
\[ = -x \cos x + \int \cos x \, dx \]
\[ = -x \cos x + \sin x \]

### Step 4: Combine the results
Now we can combine the results from Step 2 and Step 3:
\[ I = -2 \cos x - (-x \cos x + \sin x) \]
\[ = -2 \cos x + x \cos x - \sin x \]

### Step 5: Final answer
Thus, the indefinite integral is:
\[ \int (2 - x) \sin x \, dx = x \cos x - 2 \cos x - \sin x + C \]
where \( C \) is the constant of integration.