`int_(0)^(pi//2) x^(2) cos x dx`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} x^2 \cos x \, dx \), 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)
- \( dv = \cos x \, dx \) (which we will integrate)

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

Now, we find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = 2x \, dx \]
- Integrate \( dv \):
\[ v = \int \cos x \, dx = \sin x \]

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

Now we apply the integration by parts formula:
\[ I = \left[ x^2 \sin x \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \sin x \cdot 2x \, dx \]

### Step 4: Evaluate the Boundary Term

Evaluate \( \left[ x^2 \sin x \right]_{0}^{\frac{\pi}{2}} \):
- At \( x = \frac{\pi}{2} \):
\[ \left( \frac{\pi}{2} \right)^2 \sin\left( \frac{\pi}{2} \right) = \frac{\pi^2}{4} \cdot 1 = \frac{\pi^2}{4} \]
- At \( x = 0 \):
\[ 0^2 \sin(0) = 0 \]
Thus, the boundary term is:
\[ \left[ x^2 \sin x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi^2}{4} - 0 = \frac{\pi^2}{4} \]

### Step 5: Simplify the Integral

Now we need to evaluate:
\[ I = \frac{\pi^2}{4} - 2 \int_{0}^{\frac{\pi}{2}} x \sin x \, dx \]

### Step 6: Apply Integration by Parts Again

Let’s evaluate \( \int_{0}^{\frac{\pi}{2}} x \sin x \, dx \) using integration by parts again:
- Let \( u = x \) and \( dv = \sin x \, dx \)
- Then \( du = dx \) and \( v = -\cos x \)

Applying integration by parts:
\[ \int x \sin x \, dx = \left[ -x \cos x \right]_{0}^{\frac{\pi}{2}} + \int \cos x \, dx \]

### Step 7: Evaluate the New Boundary Term

Evaluate \( \left[ -x \cos x \right]_{0}^{\frac{\pi}{2}} \):
- At \( x = \frac{\pi}{2} \):
\[ -\frac{\pi}{2} \cdot \cos\left( \frac{\pi}{2} \right) = -\frac{\pi}{2} \cdot 0 = 0 \]
- At \( x = 0 \):
\[ -0 \cdot \cos(0) = 0 \]
Thus, the boundary term is:
\[ \left[ -x \cos x \right]_{0}^{\frac{\pi}{2}} = 0 - 0 = 0 \]

### Step 8: Complete the Integral

Now we have:
\[ \int_{0}^{\frac{\pi}{2}} x \sin x \, dx = 0 + \left[ -\cos x \right]_{0}^{\frac{\pi}{2}} = -\left( -1 - 1 \right) = 2 \]

### Step 9: Substitute Back into the Integral

Now substituting back:
\[ I = \frac{\pi^2}{4} - 2 \cdot 2 = \frac{\pi^2}{4} - 4 \]

### Final Result

Thus, the value of the integral is:
\[ I = \frac{\pi^2}{4} - 4 \]

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