`int sinx. Log cosx dx`

Answer

To solve the integral \( I = \int \sin x \log(\cos x) \, dx \), we can use integration by parts. Let's go through the solution step by step.

### Step 1: Set up the integral
Let:
\[ I = \int \sin x \log(\cos x) \, dx \]

### Step 2: Use substitution
We will use the substitution:
\[ t = \cos x \quad \Rightarrow \quad dt = -\sin x \, dx \quad \Rightarrow \quad dx = -\frac{dt}{\sin x} \]
Thus, we rewrite the integral:
\[ I = \int \sin x \log(\cos x) \, dx = \int \log(t) (-dt) = -\int \log(t) \, dt \]

### Step 3: Integrate using integration by parts
Now, we will apply integration by parts. Let:
- \( u = \log(t) \) \(\Rightarrow du = \frac{1}{t} dt\)
- \( dv = dt \) \(\Rightarrow v = t\)

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

### Step 4: Substitute back for \( t \)
Now substituting back \( t = \cos x \):
\[ I = -\left( \cos x \log(\cos x) - \cos x \right) = -\cos x \log(\cos x) + \cos x \]

### Step 5: Final result
Thus, the integral evaluates to:
\[ I = \cos x (1 - \log(\cos x)) + C \]
where \( C \) is the constant of integration.

### Summary of the solution:
\[ \int \sin x \log(\cos x) \, dx = \cos x (1 - \log(\cos x)) + C \]