## Step 1: Identify Odd and Even Functions A definite integral from \(-a\) to \(a\) of an **odd function** is zero, because the area on one side of the y-axis cancels out the area on the other. - **Odd function:** \(f(-x) = -f(x)\) - **Even function:** \(f(-x) = f(x)\) Let's check each integrand: 1. \(x \sin x\): \(f(-x) = (-x)\sin(-x) = -x(-\sin x) = x \sin x\) This is **even**. 2. \(\cos x\): \(f(-x) = \cos(-x) = \cos x\) This is **even**. 3. \(x \sin x \cos x\): \(f(-x) = (-x)\sin(-x)\cos(-x) = -x(-\sin x)\cos x = x \sin x \cos x\) This is **even**. 4. \(\sin^2 x \cos x\): \(f(-x) = \sin^2(-x)\cos(-x) = (\sin(-x))^2 \cos(-x) = (\!- \sin x)^2 (\cos x) = (\sin^2 x)\cos x\) This is **even**. 5. \(x^2 \sin x\): \(f(-x) = (-x)^2\sin(-x) = x^2(-\sin x) = -x^2 \sin x\) This is **odd**. **Explanation:** Only the function \(x^2 \sin x\) is odd. For definite integrals of odd functions over symmetric intervals, the result is zero. --- ## Step 2: Final Answer The only integral that evaluates to zero is: \[ \boxed{\int_{-\pi/4}^{\pi/4} x^2 \sin x \; dx} \] **Explanation:** This is because the integrand is odd, and the limits are symmetric about the origin. Thus, the positive and negative areas cancel each other out, giving a result of zero.

## Step 1: Identify Odd and Even Functions To determine which definite integrals evaluate to zero, we need to identify if the integrands are odd or even functions. Recall the definitions: - **Odd function:** \(f(-x) = -f(x)\) - **Even function:** \(f(-x) = f(x)\) We will analyze the integrands provided in the options. 1. **\( \int_{-\pi/4}^{\pi/4} \sin z \, dz \)**: - \(f(-z) = \sin(-z) = -\sin(z)\) - **This is odd.** 2. **\( \int_{-\pi/4}^{\pi/4} \cos z \, dz \)**: - \(f(-z) = \cos(-z) = \cos(z)\) - **This is even.** 3. **\( \int_{-\pi/4}^{\pi/4} z \sin z \cos z \, dz \)**: - \(f(-z) = (-z) \sin(-z) \cos(-z) = -z (-\sin z)(\cos z) = z \sin z \cos z\) - **This is odd.** 4. **\( \int_{-\pi/4}^{\pi/4} \sin^2 z \cos z \, dz \)**: - \(f(-z) = \sin^2(-z) \cos(-z) = \sin^2(z) \cos(z)\) - **This is even.** 5. **\( \int_{-\pi/4}^{\pi/4} 2 \sin z \, dz \)**: - \(f(-z) = 2 \sin(-z) = -2 \sin(z)\) - **This is odd.** ### Explanation The integrands that are odd functions are: - \( \sin z \) - \( z \sin z \cos z \) - \( 2 \sin z \) For odd functions, the definite integral over a symmetric interval around zero results in zero. --- ## Step 2: Final Answer The integrals that evaluate to zero are: 1. \(\int_{-\pi/4}^{\pi/4} \sin z \, dz\) 2. \(\int_{-\pi/4}^{\pi/4} z \sin z \cos z \, dz\) 3. \(\int_{-\pi/4}^{\pi/4} 2 \sin z \, dz\) The specific answer based on the options provided is: \[ \boxed{\int_{-\pi/4}^{\pi/4} \sin z \, dz} \] **Explanation:** All of these integrals evaluate to zero due to the properties of odd functions. However, if only one integral is to be highlighted, the integral of \(\sin z\) is a standard example of an odd function over symmetric limits resulting in zero.