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:
- \( u = x^2 \) (which we will differentiate)
- \( dv = \cos x \, dx \) (which we will integrate)

### 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

Now we can apply the integration by parts formula:

\[ \int x^2 \cos x \, dx = uv - \int v \, du \]

Substituting the values we found:

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

### Step 4: Simplify the Integral

This simplifies to:

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

### Step 5: Apply Integration by Parts Again

Now we need to evaluate \( \int x \sin x \, dx \) using integration by parts again. Let:
- \( u = x \)
- \( dv = \sin x \, dx \)

Differentiating and integrating gives us:
- \( du = dx \)
- \( v = -\cos x \)

Now applying the integration by parts formula again:

\[ \int x \sin x \, dx = uv - \int v \, du \]

Substituting the values:

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

This simplifies to:

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

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

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

### Step 6: Substitute Back

Now substitute this result back into our earlier expression:

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

This simplifies to:

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

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

### Final Answer

Thus, the evaluated integral is:

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

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