`int x^(2)sinx dx`

Answer

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

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

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

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

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

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

### Step 3: Apply the Integration by Parts Formula

Substituting into the integration by parts formula:

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

This becomes:

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

Simplifying, we have:

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

### Step 4: Solve \( \int x \cos x \, dx \)

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

Then:
- \( du = dx \)
- \( v = \sin x \)

Applying the integration by parts formula again:

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

We know that:

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

Thus:

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

### Step 5: Substitute Back

Now substituting back into our earlier equation:

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

This simplifies to:

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

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

### Final Answer

Thus, the final result of the integral is:

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

---