`int " log sin x . sec"^(2) " x dx "`

Answer

To solve the integral \( \int \log(\sin x) \sec^2 x \, dx \), we will use the integration by parts formula. The integration by parts formula is given by:

\[ \int u \, dv = uv - \int v \, du \]

### Step 1: Identify \( u \) and \( dv \)

Let:
- \( u = \log(\sin x) \)
- \( dv = \sec^2 x \, dx \)

### Step 2: Differentiate \( u \) and integrate \( dv \)

Now we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{\sin x} \cos x \, dx = \cot x \, dx \]
- Integrate \( dv \):
\[ v = \int \sec^2 x \, dx = \tan x \]

### Step 3: Apply the integration by parts formula

Now we can apply the integration by parts formula:
\[ \int \log(\sin x) \sec^2 x \, dx = \log(\sin x) \tan x - \int \tan x \cot x \, dx \]

### Step 4: Simplify the integral

Notice that:
\[ \tan x \cot x = \frac{\sin x}{\cos x} \cdot \frac{\cos x}{\sin x} = 1 \]
Thus, we have:
\[ \int \tan x \cot x \, dx = \int 1 \, dx = x \]

### Step 5: Substitute back into the equation

Now substituting back, we get:
\[ \int \log(\sin x) \sec^2 x \, dx = \log(\sin x) \tan x - x + C \]

### Final Answer

The final result is:
\[ \int \log(\sin x) \sec^2 x \, dx = \log(\sin x) \tan x - x + C \]