`intsin2xlog_(e)cosx dx` is equal to

Options

A. `((1)/(2)+log_(e)cosx)cos^(2)x+C`
B. `cos^(2)x.log_(e)cosx+C`
C. `((1)/(2)+log_(e)cosx)cos^(2)x+C`
D. none of these

Explanation

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

### Step 1: Rewrite \( \sin(2x) \)
Using the double angle identity for sine, we have:
\[ \sin(2x) = 2 \sin(x) \cos(x) \]
Thus, we can rewrite the integral as:
\[ \int \sin(2x) \log(\cos x) \, dx = \int 2 \sin(x) \cos(x) \log(\cos x) \, dx \]

### Step 2: Substitution
Let \( t = \cos(x) \). Then, the derivative \( dt = -\sin(x) \, dx \) or \( \sin(x) \, dx = -dt \). The integral becomes:
\[ \int 2 \sin(x) \cos(x) \log(\cos x) \, dx = -2 \int t \log(t) \, dt \]

### Step 3: Integration by Parts
Now we will use integration by parts to solve \( -2 \int t \log(t) \, dt \). Let:
- \( u = \log(t) \) ⇒ \( du = \frac{1}{t} \, dt \)
- \( dv = t \, dt \) ⇒ \( v = \frac{t^2}{2} \)

Applying integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Substituting the values:
\[ -2 \left( \log(t) \cdot \frac{t^2}{2} - \int \frac{t^2}{2} \cdot \frac{1}{t} \, dt \right) \]
This simplifies to:
\[ -2 \left( \frac{t^2}{2} \log(t) - \int \frac{t}{2} \, dt \right) \]

### Step 4: Solve the Remaining Integral
The remaining integral is:
\[ \int \frac{t}{2} \, dt = \frac{t^2}{4} \]
Thus, we have:
\[ -2 \left( \frac{t^2}{2} \log(t) - \frac{t^2}{4} \right) = -t^2 \log(t) + \frac{t^2}{2} \]

### Step 5: Substitute Back
Now substitute \( t = \cos(x) \):
\[ -t^2 \log(t) + \frac{t^2}{2} = -\cos^2(x) \log(\cos(x)) + \frac{\cos^2(x)}{2} \]

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