`int sin 2x. Logcos x dx is ` equal to

Options

A. `cos ^(2) x ((1)/(2) + log cos x) +k`
B. `cos ^(2) x.log cos x +k`
C. `cos ^(2) x ((1)/(2) - log cos x ) +k`
D. None of these

Answer

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

### Step 1: Rewrite the Integral
We know that \( \sin(2x) = 2 \sin(x) \cos(x) \). Therefore, we can rewrite the integral as:
\[ I = \int 2 \sin(x) \cos(x) \log(\cos x) \, dx \]

### Step 2: Use Substitution
Let \( t = \cos(x) \). Then, the derivative of \( t \) with respect to \( x \) is:
\[ dt = -\sin(x) \, dx \quad \Rightarrow \quad dx = -\frac{dt}{\sin(x)} \]
Now, substituting \( \sin(x) = \sqrt{1 - t^2} \) (since \( \sin^2(x) + \cos^2(x) = 1 \)), we have:
\[ I = \int 2 \sin(x) t \log(t) \left(-\frac{dt}{\sin(x)}\right) \]
This simplifies to:
\[ I = -2 \int t \log(t) \, dt \]

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

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get:
\[ \int t \log(t) \, dt = \frac{t^2}{2} \log(t) - \int \frac{t^2}{2} \cdot \frac{1}{t} \, dt \]
This simplifies to:
\[ \int t \log(t) \, dt = \frac{t^2}{2} \log(t) - \frac{1}{2} \int t \, dt \]
Calculating the remaining integral:
\[ \int t \, dt = \frac{t^2}{2} \]
Thus, we have:
\[ \int t \log(t) \, dt = \frac{t^2}{2} \log(t) - \frac{1}{4} t^2 + C \]

### Step 4: Substitute Back
Now substituting back into our expression for \( I \):
\[ I = -2 \left( \frac{t^2}{2} \log(t) - \frac{1}{4} t^2 \right) + C \]
This simplifies to:
\[ I = -t^2 \log(t) + \frac{1}{2} t^2 + C \]

### Step 5: Substitute \( t = \cos(x) \)
Finally, substituting \( t = \cos(x) \) back into the equation:
\[ I = -\cos^2(x) \log(\cos(x)) + \frac{1}{2} \cos^2(x) + C \]

### Final Answer
Thus, the integral evaluates to:
\[ I = \frac{1}{2} \cos^2(x) - \cos^2(x) \log(\cos(x)) + C \]