`int x^(2) "arc" cos x dx`.

Answer

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

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

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

Let:
- \( u = \arccos(x) \) (this is the inverse trigonometric function)
- \( dv = x^2 \, dx \) (this is the algebraic function)

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

Now we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = -\frac{1}{\sqrt{1 - x^2}} \, dx \]

- Integrate \( dv \):
\[ v = \int x^2 \, dx = \frac{x^3}{3} \]

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

Now we can apply the integration by parts formula:
\[ \int x^2 \arccos(x) \, dx = uv - \int v \, du \]
Substituting the values of \( u \), \( v \), and \( du \):
\[ = \arccos(x) \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \left(-\frac{1}{\sqrt{1 - x^2}}\right) \, dx \]
This simplifies to:
\[ = \frac{x^3}{3} \arccos(x) + \frac{1}{3} \int \frac{x^3}{\sqrt{1 - x^2}} \, dx \]

### Step 4: Simplify the remaining integral

To solve \( \int \frac{x^3}{\sqrt{1 - x^2}} \, dx \), we can use the substitution:
Let \( x = \sin(\theta) \), then \( dx = \cos(\theta) \, d\theta \) and \( \sqrt{1 - x^2} = \cos(\theta) \).

Substituting these into the integral gives:
\[ \int \frac{\sin^3(\theta)}{\cos(\theta)} \cos(\theta) \, d\theta = \int \sin^3(\theta) \, d\theta \]

### Step 5: Solve \( \int \sin^3(\theta) \, d\theta \)

Using the identity \( \sin^3(\theta) = \sin(\theta)(1 - \cos^2(\theta)) \):
\[ = \int \sin(\theta) \, d\theta - \int \sin(\theta) \cos^2(\theta) \, d\theta \]

The first integral is:
\[ -\cos(\theta) \]

For the second integral, use the substitution \( u = \cos(\theta) \), \( du = -\sin(\theta) d\theta \):
\[ \int \sin(\theta) \cos^2(\theta) \, d\theta = -\int u^2 \, du = -\frac{u^3}{3} = -\frac{\cos^3(\theta)}{3} \]

Thus,
\[ \int \sin^3(\theta) \, d\theta = -\cos(\theta) + \frac{\cos^3(\theta)}{3} \]

### Step 6: Substitute back to \( x \)

Now substitute back \( \theta = \arcsin(x) \):
\[ = -\sqrt{1 - x^2} + \frac{(1 - x^2)^{3/2}}{3} \]

### Final Step: Combine everything

Putting everything together:
\[ \int x^2 \arccos(x) \, dx = \frac{x^3}{3} \arccos(x) + \frac{1}{3} \left(-\sqrt{1 - x^2} + \frac{(1 - x^2)^{3/2}}{3}\right) + C \]

Thus, the final answer is:
\[ \int x^2 \arccos(x) \, dx = \frac{x^3}{3} \arccos(x) - \frac{\sqrt{1 - x^2}}{3} + \frac{(1 - x^2)^{3/2}}{9} + C \]