`int(e^(logx)+sinx).cos x dx=`

Options

A. `x.cos x-sin x+4.cos 2x+c`
B. `x.sinx+cos x-(1)/(4).cos 2x+c`
C. `x.secx +tanx-(1)/(2).sin2x+c`
D. none of these

Answer

To solve the integral \( I = \int (e^{\log x} + \sin x) \cos x \, dx \), we can follow these steps:

### Step 1: Simplify \( e^{\log x} \)
Using the property of logarithms, we know that:
\[ e^{\log x} = x \]
Thus, we can rewrite the integral as:
\[ I = \int (x + \sin x) \cos x \, dx \]

### Step 2: Distribute \( \cos x \)
Now, we can distribute \( \cos x \) across the terms in the integral:
\[ I = \int x \cos x \, dx + \int \sin x \cos x \, dx \]

### Step 3: Solve \( \int x \cos x \, dx \)
To solve \( \int x \cos x \, dx \), we can use integration by parts. Let:
- \( u = x \) and \( dv = \cos x \, dx \)
Then, we have:
- \( du = dx \) and \( v = \sin x \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int x \cos x \, dx = x \sin x - \int \sin x \, dx \]
Now, we compute \( \int \sin x \, dx \):
\[ \int \sin x \, dx = -\cos x \]
Thus,
\[ \int x \cos x \, dx = x \sin x + \cos x \]

### Step 4: Solve \( \int \sin x \cos x \, dx \)
We can use the double angle identity for sine:
\[ \sin x \cos x = \frac{1}{2} \sin(2x) \]
Thus,
\[ \int \sin x \cos x \, dx = \int \frac{1}{2} \sin(2x) \, dx \]
Now, integrating \( \frac{1}{2} \sin(2x) \):
\[ \int \frac{1}{2} \sin(2x) \, dx = -\frac{1}{4} \cos(2x) \]

### Step 5: Combine the results
Now, we can combine the results of both integrals:
\[ I = \left( x \sin x + \cos x \right) - \frac{1}{4} \cos(2x) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ I = x \sin x + \cos x - \frac{1}{4} \cos(2x) + C \]