`int cos(logx)dx=?`

Options

A. `(x)/(2)[cos(logx)-sin(logx)]+C`
B. `(x)/(2)[cos(logx)+sin(logx)]+C`
C. `2x[cos(logx)+sin(logx)]+C`
D. `2x[cos(logx)-sin(logx)]+C`

Explanation

To solve the integral \( \int \cos(\log x) \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = \log x \). Then, we can express \( x \) in terms of \( t \):
\[ x = e^t \]
Now, we need to find \( dx \):
\[ dx = e^t \, dt \]

### Step 2: Rewrite the Integral
Substituting \( t \) and \( dx \) into the integral, we have:
\[ \int \cos(\log x) \, dx = \int \cos(t) \cdot e^t \, dt \]

### Step 3: Integration by Parts
Now we will use integration by parts. Let:
- \( u = e^t \) (first function)
- \( dv = \cos(t) \, dt \) (second function)

Then we need to find \( du \) and \( v \):
\[ du = e^t \, dt, \quad v = \sin(t) \]

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int e^t \cos(t) \, dt = e^t \sin(t) - \int \sin(t) e^t \, dt \]

### Step 4: Apply Integration by Parts Again
Now we need to integrate \( \int \sin(t) e^t \, dt \) using integration by parts again. Let:
- \( u = e^t \)
- \( dv = \sin(t) \, dt \)

Then:
\[ du = e^t \, dt, \quad v = -\cos(t) \]

So,
\[ \int e^t \sin(t) \, dt = -e^t \cos(t) - \int -\cos(t) e^t \, dt \]
\[ = -e^t \cos(t) + \int e^t \cos(t) \, dt \]

### Step 5: Combine the Results
Now, substituting back into our equation:
\[ \int e^t \cos(t) \, dt = e^t \sin(t) - \left( -e^t \cos(t) + \int e^t \cos(t) \, dt \right) \]
\[ = e^t \sin(t) + e^t \cos(t) - \int e^t \cos(t) \, dt \]

Let \( I = \int e^t \cos(t) \, dt \). Then we have:
\[ I = e^t \sin(t) + e^t \cos(t) - I \]
\[ 2I = e^t (\sin(t) + \cos(t)) \]
\[ I = \frac{e^t}{2} (\sin(t) + \cos(t)) \]

### Step 6: Substitute Back
Now substituting back \( t = \log x \):
\[ I = \frac{e^{\log x}}{2} (\sin(\log x) + \cos(\log x)) = \frac{x}{2} (\sin(\log x) + \cos(\log x)) \]

### Step 7: Final Answer
Thus, the final answer for the integral is:
\[ \int \cos(\log x) \, dx = \frac{x}{2} (\sin(\log x) + \cos(\log x)) + C \]