What is `int(xcosx+sinx)dx` equal to

Options

A. `xsinx+C`
B. `xcosx+C`
C. `-xsinx+C`
D. `-xcosx+C`

Answer

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 \( \int x \cos x \, dx \) using Integration by Parts

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

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

Let:
- \( u = x \) (which is an algebraic function)
- \( dv = \cos x \, dx \)

Now, we need to find \( du \) and \( v \):
- \( du = dx \)
- \( v = \int \cos x \, dx = \sin x \)

Now, applying the integration by parts formula:

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

### Step 2: Solve \( \int \sin x \, dx \)

Now we need to solve \( \int \sin x \, dx \):

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

### Step 3: Combine the results

Substituting back into our equation from Step 1:

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

### Step 4: Combine with the second integral

Now, we can combine this with the second integral we initially separated:

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

The \( \cos x \) terms cancel out:

\[ = x \sin x + C \]

Where \( C \) is the constant of integration.

### Final Answer

Thus, the final result is:

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