`int(cos^(2) (log .x))/(x)dx`

Answer

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

### Step 1: Substitution
Let \( t = \log x \). Then, the differential \( dx \) can be expressed in terms of \( dt \):
\[ dx = e^t \, dt = x \, dt \]
Thus, we can rewrite the integral as:
\[ \int \frac{\cos^2(t)}{e^t} \, e^t \, dt = \int \cos^2(t) \, dt \]

### Step 2: Use the Cosine Double Angle Identity
We know from trigonometric identities that:
\[ \cos^2(t) = \frac{1 + \cos(2t)}{2} \]
Substituting this into the integral gives:
\[ \int \cos^2(t) \, dt = \int \frac{1 + \cos(2t)}{2} \, dt \]

### Step 3: Split the Integral
We can split the integral into two parts:
\[ \int \frac{1 + \cos(2t)}{2} \, dt = \frac{1}{2} \int 1 \, dt + \frac{1}{2} \int \cos(2t) \, dt \]

### Step 4: Integrate Each Part
Now, we integrate each part:
1. The integral of \( 1 \) is:
\[ \int 1 \, dt = t \]
2. The integral of \( \cos(2t) \) is:
\[ \int \cos(2t) \, dt = \frac{\sin(2t)}{2} \]

Putting it all together, we have:
\[ \frac{1}{2} t + \frac{1}{2} \cdot \frac{\sin(2t)}{2} + C = \frac{1}{2} t + \frac{\sin(2t)}{4} + C \]

### Step 5: Substitute Back for \( t \)
Since we let \( t = \log x \), we substitute back:
\[ \frac{1}{2} \log x + \frac{\sin(2 \log x)}{4} + C \]

### Final Answer
Thus, the final answer for the integral is:
\[ \int \frac{\cos^2(\log x)}{x} \, dx = \frac{1}{2} \log x + \frac{\sin(2 \log x)}{4} + C \]