`int xcos^(2)xdx=`

Options

A. `x^(2)/4-1/4xsin2x-1/8cos2x+c`
B. `x^(2)/4+1/4xsin2x+1/8cos2x+c`
C. `x^(4)-1/4xsin2x+1/8cos2x+c`
D. `x^(4)+1/4xsin2x-1/8cos2x+c`

Answer

To solve the integral \( \int x \cos^2 x \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
Using the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \), we can rewrite the integral:
\[ \int x \cos^2 x \, dx = \int x \cdot \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2} \int x (1 + \cos 2x) \, dx \]

### Step 2: Split the integral
Now, we can split the integral into two parts:
\[ \frac{1}{2} \int x (1 + \cos 2x) \, dx = \frac{1}{2} \left( \int x \, dx + \int x \cos 2x \, dx \right) \]

### Step 3: Solve the first integral
The first integral is straightforward:
\[ \int x \, dx = \frac{x^2}{2} \]

### Step 4: Solve the second integral using integration by parts
For the second integral \( \int x \cos 2x \, dx \), we will use integration by parts. Let:
- \( u = x \) (then \( du = dx \))
- \( dv = \cos 2x \, dx \) (then \( v = \frac{1}{2} \sin 2x \))

Applying integration by parts:
\[ \int x \cos 2x \, dx = uv - \int v \, du = x \cdot \frac{1}{2} \sin 2x - \int \frac{1}{2} \sin 2x \, dx \]

### Step 5: Solve the integral \( \int \sin 2x \, dx \)
The integral \( \int \sin 2x \, dx \) can be solved as follows:
\[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x \]

### Step 6: Substitute back
Now substituting back into our integration by parts result:
\[ \int x \cos 2x \, dx = \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x \]

### Step 7: Combine all parts
Now we can combine all parts:
\[ \frac{1}{2} \left( \int x \, dx + \int x \cos 2x \, dx \right) = \frac{1}{2} \left( \frac{x^2}{2} + \left( \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x \right) \right) \]

### Step 8: Final expression
Putting everything together, we have:
\[ \int x \cos^2 x \, dx = \frac{1}{4} x^2 + \frac{1}{4} x \sin 2x + \frac{1}{8} \cos 2x + C \]

Where \( C \) is the constant of integration.

### Final Answer:
\[ \int x \cos^2 x \, dx = \frac{x^2}{4} + \frac{x \sin 2x}{4} + \frac{\cos 2x}{8} + C \]

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