The value of `int_(-pi)^(pi) (1-x^(2)) sin x cos^(2) x dx` is

Options

A. 0
B. `pi- (pi^(3))/(3)`
C. `2pi- pi^(3)`
D. `(pi)/(2)- 2pi^(3)`

Answer

To solve the integral \( \int_{-\pi}^{\pi} (1 - x^2) \sin x \cos^2 x \, dx \), we can use the property of definite integrals that states if \( f(-x) = -f(x) \), then \( \int_{-a}^{a} f(x) \, dx = 0 \).

### Step-by-Step Solution:

1. **Define the function**:
Let \( f(x) = (1 - x^2) \sin x \cos^2 x \).

2. **Find \( f(-x) \)**:
We need to compute \( f(-x) \):
\[ f(-x) = (1 - (-x)^2) \sin(-x) \cos^2(-x) \]
Simplifying this:
- \( (-x)^2 = x^2 \), so \( 1 - (-x)^2 = 1 - x^2 \).
- \( \sin(-x) = -\sin x \).
- \( \cos(-x) = \cos x \), so \( \cos^2(-x) = \cos^2 x \).

Therefore,
\[ f(-x) = (1 - x^2)(-\sin x)(\cos^2 x) = -(1 - x^2) \sin x \cos^2 x = -f(x). \]

3. **Apply the property of definite integrals**:
Since we have shown that \( f(-x) = -f(x) \), we can use the property:
\[ \int_{-a}^{a} f(x) \, dx = 0. \]
Thus,
\[ \int_{-\pi}^{\pi} (1 - x^2) \sin x \cos^2 x \, dx = 0. \]

### Final Answer:
The value of the integral is \( 0 \).