`int{sin(log_(e)x)+cos(log_(e)x)}dx` is equal to

Options

A. `sin(log_(e)x)+cos(log_(e)x)+C`
B. `xsin (log_(e)x)+C`
C. `xcos(log_(e)x)+C`
D. none of these

Explanation

To solve the integral \( I = \int \left( \sin(\log_e x) + \cos(\log_e x) \right) dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = \log_e x \). Then, we differentiate both sides:
\[ dt = \frac{1}{x} dx \quad \Rightarrow \quad dx = x \, dt \]
Since \( x = e^t \), we can substitute this into our expression for \( dx \):
\[ dx = e^t \, dt \]

### Step 2: Rewrite the Integral
Now, we can rewrite the integral \( I \) in terms of \( t \):
\[ I = \int \left( \sin(t) + \cos(t) \right) e^t \, dt \]

### Step 3: Distribute the Integral
We can separate the integral:
\[ I = \int \sin(t) e^t \, dt + \int \cos(t) e^t \, dt \]

### Step 4: Use Integration by Parts
For both integrals, we will use integration by parts. Recall the formula:
\[ \int u \, dv = uv - \int v \, du \]

#### Integral 1: \( \int \sin(t) e^t \, dt \)
Let \( u = \sin(t) \) and \( dv = e^t dt \).
Then, \( du = \cos(t) dt \) and \( v = e^t \).

Applying integration by parts:
\[ \int \sin(t) e^t \, dt = \sin(t) e^t - \int e^t \cos(t) \, dt \]

#### Integral 2: \( \int \cos(t) e^t \, dt \)
Let \( u = \cos(t) \) and \( dv = e^t dt \).
Then, \( du = -\sin(t) dt \) and \( v = e^t \).

Applying integration by parts:
\[ \int \cos(t) e^t \, dt = \cos(t) e^t + \int e^t \sin(t) \, dt \]

### Step 5: Combine Results
Now we have:
\[ I = \left( \sin(t) e^t - \int e^t \cos(t) \, dt \right) + \left( \cos(t) e^t + \int e^t \sin(t) \, dt \right) \]
This simplifies to:
\[ I = \sin(t) e^t + \cos(t) e^t - \int e^t \cos(t) \, dt + \int e^t \sin(t) \, dt \]

### Step 6: Solve for \( I \)
Let \( J = \int e^t \cos(t) \, dt \). Then we can express:
\[ I = e^t (\sin(t) + \cos(t)) - J + I \]
Rearranging gives:
\[ 2I = e^t (\sin(t) + \cos(t)) - J \]
Now, we can substitute \( J \) back into the equation and solve for \( I \).

### Step 7: Final Result
After solving and substituting back \( t = \log_e x \):
\[ I = x (\sin(\log_e x) + \cos(\log_e x)) + C \]
Thus, the final answer is:
\[ I = x \sin(\log_e x) + x \cos(\log_e x) + C \]