The value of `int(x e^ln(sinx)-cosx)dx` is equal to

Options

A. `xcosx+C`
B. `sinx-xcosx+C`
C. `-e^(ln x)cosx+C`
D. `sinx+xcosx+C`

Answer

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

### Step 1: Simplify the expression
We know that \( e^{\ln(a)} = a \). Therefore, we can simplify \( e^{\ln(\sin x)} \) to \( \sin x \). This gives us:
\[ I = \int (x \sin x - \cos x) \, dx \]

### Step 2: Separate the integral
We can separate the integral into two parts:
\[ I = \int x \sin x \, dx - \int \cos x \, dx \]

### Step 3: Solve the integral \( \int x \sin x \, dx \) using integration by parts
Let:
- \( u = x \) (then \( du = dx \))
- \( dv = \sin x \, dx \) (then \( v = -\cos x \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int x \sin x \, dx = -x \cos x - \int -\cos x \, dx \]
This simplifies to:
\[ \int x \sin x \, dx = -x \cos x + \int \cos x \, dx \]

### Step 4: Solve the integral \( \int \cos x \, dx \)
The integral of \( \cos x \) is:
\[ \int \cos x \, dx = \sin x \]

### Step 5: Combine the results
Substituting back into our expression for \( \int x \sin x \, dx \):
\[ \int x \sin x \, dx = -x \cos x + \sin x \]

### Step 6: Substitute back into the original integral
Now substituting this back into our expression for \( I \):
\[ I = (-x \cos x + \sin x) - \int \cos x \, dx \]
Since \( \int \cos x \, dx = \sin x \), we have:
\[ I = -x \cos x + \sin x - \sin x \]
The \( \sin x \) terms cancel out:
\[ I = -x \cos x + C \]

### Final Result
Thus, the value of the integral is:
\[ I = -x \cos x + C \]