Evaluate `int x^(2) cos x dx`

Answer

To evaluate the integral \( \int x^2 \cos x \, dx \), we will use the method of integration by parts. The integration by parts formula is given by:

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

### Step 1: Choose \( u \) and \( dv \)

Let:
- \( u = x^2 \) (first function)
- \( dv = \cos x \, dx \) (second function)

### Step 2: Differentiate \( u \) and integrate \( dv \)

Now we differentiate \( u \) and integrate \( dv \):
- \( du = 2x \, dx \)
- \( v = \int \cos x \, dx = \sin x \)

### Step 3: Apply the integration by parts formula

Substituting into the integration by parts formula:

\[ \int x^2 \cos x \, dx = x^2 \sin x - \int \sin x \cdot 2x \, dx \]

### Step 4: Simplify the integral

Now we need to evaluate \( \int 2x \sin x \, dx \). We will apply integration by parts again.

Let:
- \( u = 2x \)
- \( dv = \sin x \, dx \)

Then,
- \( du = 2 \, dx \)
- \( v = -\cos x \)

### Step 5: Apply integration by parts again

Using the integration by parts formula again:

\[ \int 2x \sin x \, dx = 2x (-\cos x) - \int -\cos x \cdot 2 \, dx \]
\[ = -2x \cos x + 2 \int \cos x \, dx \]
\[ = -2x \cos x + 2 \sin x \]

### Step 6: Substitute back into the original integral

Now substitute this result back into our earlier expression:

\[ \int x^2 \cos x \, dx = x^2 \sin x - (-2x \cos x + 2 \sin x) \]
\[ = x^2 \sin x + 2x \cos x - 2 \sin x + C \]

### Final Result

Thus, the final result of the integral is:

\[ \int x^2 \cos x \, dx = x^2 \sin x + 2x \cos x - 2 \sin x + C \]