`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 integration by parts. The integration by parts formula is:

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

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

Let:
- \( u = x^2 \) (so that \( du = 2x \, dx \))
- \( dv = \cos x \, dx \) (so that \( v = \sin x \))

### Step 2: Apply Integration by Parts

Using the integration by parts formula:

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

Substituting the values we have:

\[ I = x^2 \sin x \Big|_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \sin x \cdot 2x \, dx \]

### Step 3: Evaluate the Boundary Term

Now evaluate \( x^2 \sin x \) at the limits \( 0 \) and \( \frac{\pi}{2} \):

\[ x^2 \sin x \Big|_{0}^{\frac{\pi}{2}} = \left( \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) \right) - \left( 0^2 \sin(0) \right) = \left( \frac{\pi^2}{4} \cdot 1 \right) - 0 = \frac{\pi^2}{4} \]

### Step 4: Simplify the Remaining Integral

Now we need to compute the integral \( \int_{0}^{\frac{\pi}{2}} 2x \sin x \, dx \). We will apply integration by parts again.

Let:
- \( u = 2x \) (so that \( du = 2 \, dx \))
- \( dv = \sin x \, dx \) (so that \( v = -\cos x \))

Applying integration by parts again:

\[ \int 2x \sin x \, dx = 2x (-\cos x) \Big|_{0}^{\frac{\pi}{2}} - \int -\cos x \cdot 2 \, dx \]

### Step 5: Evaluate the Boundary Term

Evaluate \( 2x (-\cos x) \) at the limits \( 0 \) and \( \frac{\pi}{2} \):

\[ 2x (-\cos x) \Big|_{0}^{\frac{\pi}{2}} = \left( 2\left(\frac{\pi}{2}\right)(-\cos\left(\frac{\pi}{2}\right)) \right) - \left( 2(0)(-\cos(0)) \right) = 0 - 0 = 0 \]

### Step 6: Calculate the Remaining Integral

Now we compute the remaining integral:

\[ \int -2 \cos x \, dx = -2 \sin x \Big|_{0}^{\frac{\pi}{2}} = -2(1 - 0) = -2 \]

### Step 7: Combine Results

Now substitute back into the equation for \( I \):

\[ I = \frac{\pi^2}{4} - 0 + 2 = \frac{\pi^2}{4} + 2 \]

### Final Answer

Thus, the value of the integral is:

\[ \int_{0}^{\frac{\pi}{2}} x^2 \cos x \, dx = \frac{\pi^2}{4} + 2 \]

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