Question

$\int_{-\pi}^\pi (1 - x^2)\sin x \cos^2 x \, dx =$

• A. $\frac{\pi}{3}$
• B. $2\pi - \pi^2$
• C. $\frac{\pi^3}{2}$
• D. $0$

Answer 1

The Correct Option is D

1. Understand the integral:

We need to evaluate \( \int_{-\pi}^{\pi} (1 - x^2) \sin x \cdot \cos^2 x \, dx \).

2. Analyze the integrand:

The integrand is \( (1 - x^2) \sin x \cos^2 x \). Notice that:

• \( (1 - x^2) \) is an even function.
• \( \sin x \) is an odd function.
• \( \cos^2 x \) is an even function.

The product of an even function and an odd function is odd, so \( (1 - x^2) \sin x \cos^2 x \) is odd.

3. Use the property of odd functions:

For an odd function \( f(x) \), the integral over symmetric limits \([-a, a]\) is zero:

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

Thus, the given integral evaluates to 0.

Correct Answer: (D) 0

Answer 2

The integral is 0 because the integrand is an odd function, and we're integrating over a symmetric interval around 0.

• $ \sin(x) $ is an odd function.
• $ \cos^2(x) $ is an even function.
• $ (1 - x^2) $ is an even function.

The product of two even functions is even, and the product of an even and an odd function is odd. 

Thus, $ (1 - x^2)\sin(x)\cos^2(x) $ is an odd function.

The integral of an odd function over a symmetric interval $[-a, a]$ is always 0.

Therefore,

\[ \int_{-\pi}^{\pi} (1 - x^2)\sin(x)\cos^2(x) \, dx = 0 \]