What is `int (x cos x + sin x) dx` equal to ?
Where c is an arbitrary constant

Options

A. `x sin x + c`
B. `x cos x + c`
C. `-x sin x + c`
D. `-x cos x + c`

Explanation

To solve the integral \( \int (x \cos x + \sin x) \, dx \), we can break it down into two separate integrals:

\[ \int (x \cos x + \sin x) \, dx = \int x \cos x \, dx + \int \sin x \, dx \]

### Step 1: Solve \( \int \sin x \, dx \)

The integral of \( \sin x \) is straightforward:

\[ \int \sin x \, dx = -\cos x + C_1 \]

### Step 2: Solve \( \int x \cos x \, dx \) using Integration by Parts

For the integral \( \int x \cos x \, dx \), we will use the integration by parts formula:

\[ \int u \, dv = uv - \int v \, du \]

Here, we can choose:
- \( u = x \) (thus \( du = dx \))
- \( dv = \cos x \, dx \) (thus \( v = \sin x \))

Now, applying the integration by parts:

\[ \int x \cos x \, dx = x \sin x - \int \sin x \, dx \]

### Step 3: Substitute \( \int \sin x \, dx \)

Now we substitute \( \int \sin x \, dx \) back into our equation:

\[ \int x \cos x \, dx = x \sin x - (-\cos x + C_1) \]

This simplifies to:

\[ \int x \cos x \, dx = x \sin x + \cos x - C_1 \]

### Step 4: Combine the results

Now we can combine the results of both integrals:

\[ \int (x \cos x + \sin x) \, dx = \left( x \sin x + \cos x - C_1 \right) + (-\cos x + C_1) \]

The \( \cos x \) terms cancel out, and we are left with:

\[ \int (x \cos x + \sin x) \, dx = x \sin x + C \]

where \( C \) is an arbitrary constant.

### Final Answer:

\[ \int (x \cos x + \sin x) \, dx = x \sin x + C \]

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