Evaluate: `intsin^2(logx)dx`

Explanation

To evaluate the integral \( I = \int \sin^2(\log x) \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = \log x \). Then, we have:
\[ x = e^t \quad \text{and} \quad dx = e^t \, dt \]
Substituting these into the integral gives:
\[ I = \int \sin^2(t) \cdot e^t \, dt \]

### Step 2: Use the identity for \(\sin^2(t)\)
We can use the identity:
\[ \sin^2(t) = \frac{1 - \cos(2t)}{2} \]
Thus, we can rewrite the integral as:
\[ I = \int e^t \cdot \frac{1 - \cos(2t)}{2} \, dt = \frac{1}{2} \int e^t \, dt - \frac{1}{2} \int e^t \cos(2t) \, dt \]

### Step 3: Evaluate the first integral
The first integral is straightforward:
\[ \int e^t \, dt = e^t \]
So,
\[ \frac{1}{2} \int e^t \, dt = \frac{1}{2} e^t \]

### Step 4: Evaluate the second integral using integration by parts
Let:
\[ I_1 = \int e^t \cos(2t) \, dt \]
Using integration by parts, let \( u = \cos(2t) \) and \( dv = e^t \, dt \). Then:
\[ du = -2 \sin(2t) \, dt \quad \text{and} \quad v = e^t \]
Thus, we have:
\[ I_1 = e^t \cos(2t) - \int e^t (-2 \sin(2t)) \, dt \]
This simplifies to:
\[ I_1 = e^t \cos(2t) + 2 \int e^t \sin(2t) \, dt \]
Let \( I_2 = \int e^t \sin(2t) \, dt \). We will also use integration by parts on \( I_2 \):
Let \( u = \sin(2t) \) and \( dv = e^t \, dt \):
\[ du = 2 \cos(2t) \, dt \quad \text{and} \quad v = e^t \]
Thus,
\[ I_2 = e^t \sin(2t) - \int e^t (2 \cos(2t)) \, dt \]
This gives:
\[ I_2 = e^t \sin(2t) - 2 I_1 \]

### Step 5: Solve the system of equations
Now we have two equations:
1. \( I_1 = e^t \cos(2t) + 2 I_2 \)
2. \( I_2 = e^t \sin(2t) - 2 I_1 \)

Substituting equation 2 into equation 1:
\[ I_1 = e^t \cos(2t) + 2(e^t \sin(2t) - 2 I_1) \]
This simplifies to:
\[ I_1 + 4 I_1 = e^t \cos(2t) + 2 e^t \sin(2t) \]
Thus:
\[ 5 I_1 = e^t \cos(2t) + 2 e^t \sin(2t) \]
So,
\[ I_1 = \frac{1}{5}(e^t \cos(2t) + 2 e^t \sin(2t)) \]

### Step 6: Substitute back to original variable
Now substituting back \( t = \log x \):
\[ I_1 = \frac{1}{5}(x \cos(2 \log x) + 2x \sin(2 \log x)) \]
Thus, our integral becomes:
\[ I = \frac{1}{2} e^t - \frac{1}{10}(e^t \cos(2t) + 2 e^t \sin(2t)) + C \]
Substituting back \( e^t = x \):
\[ I = \frac{x}{2} - \frac{1}{10}(x \cos(2 \log x) + 2x \sin(2 \log x)) + C \]

### Final Answer
\[ I = \frac{x}{2} - \frac{1}{10} x \cos(2 \log x) + \frac{1}{5} x \sin(2 \log x) + C \]