Evaluate `int_(-pi/2)^(pi/2) x^2 sin x dx `

Answer

To evaluate the integral

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \sin x \, dx, \]

we will use the properties of odd and even functions.

### Step 1: Identify the function
Let

\[ f(x) = x^2 \sin x. \]

### Step 2: Check if the function is odd or even
To determine if \( f(x) \) is odd or even, we need to evaluate \( f(-x) \):

\[ f(-x) = (-x)^2 \sin(-x) = x^2 (-\sin x) = -x^2 \sin x = -f(x). \]

Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is an odd function.

### Step 3: Apply the property of definite integrals
The property of definite integrals states that if \( f(x) \) is an odd function, then:

\[ \int_{-a}^{a} f(x) \, dx = 0. \]

In our case, since \( f(x) = x^2 \sin x \) is odd and we are integrating from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), we have:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \sin x \, dx = 0. \]

### Final Answer
Thus, the value of the integral is

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \sin x \, dx = 0. \]