Evaluate ∫xcos(2x)dx by integration by parts method ?

To solve the integral ∫xcos(2x)dx using the integration by parts method, follow these steps:

### Formula for Integration by Parts:
The formula for integration by parts is:
∫u dv = uv - ∫v du

Here, we need to choose \( u \) and \( dv \) from \( x \) and \( \cos(2x)dx \). A general rule is to choose \( u \) as the function that simplifies upon differentiation. So, let:
- \( u = x \) (it simplifies to 1 upon differentiation)
- \( dv = \cos(2x)dx \) (this integrates easily)

### Step 1: Differentiate \( u \) and integrate \( dv \)
- \( u = x \) → \( du = dx \)
- \( dv = \cos(2x)dx \) → \( v = \frac{\sin(2x)}{2} \)

### Step 2: Apply the integration by parts formula
Substitute into the formula:
∫xcos(2x)dx = uv - ∫v du

Using \( u = x \), \( v = \frac{\sin(2x)}{2} \), and \( du = dx \):
∫xcos(2x)dx = \( x \cdot \frac{\sin(2x)}{2} \) - ∫\( \frac{\sin(2x)}{2}dx \)

### Step 3: Simplify the first term and evaluate the remaining integral
∫xcos(2x)dx = \( \frac{x\sin(2x)}{2} \) - \( \frac{1}{2} \int\sin(2x)dx \)

Now, calculate \( \int\sin(2x)dx \):
- The integral of \( \sin(2x) \) is \( -\frac{\cos(2x)}{2} \).

Substitute this back:
∫xcos(2x)dx = \( \frac{x\sin(2x)}{2} \) - \( \frac{1}{2} \cdot \left( -\frac{\cos(2x)}{2} \right) \)

### Step 4: Simplify the expression
∫xcos(2x)dx = \( \frac{x\sin(2x)}{2} \) + \( \frac{\cos(2x)}{4} \)

### Step 5: Add the constant of integration
The final answer is:
∫xcos(2x)dx = \( \frac{x\sin(2x)}{2} + \frac{\cos(2x)}{4} + C \), where \( C \) is the constant of integration.