Question

(c) Find the value of \( \int_{-\pi/2}^{\pi/2} \sin^2 x \, dx \):

Hint:

Use trigonometric identities like \( \sin^2 x = \frac{1 - \cos 2x}{2} \) to simplify integrals.

Answer 1

Using the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \): \[ \int_{-\pi/2}^{\pi/2} \sin^2 x \, dx = \int_{-\pi/2}^{\pi/2} \frac{1 - \cos 2x}{2} \, dx. \] Separate the terms: \[ \int_{-\pi/2}^{\pi/2} \sin^2 x \, dx = \frac{1}{2} \int_{-\pi/2}^{\pi/2} 1 \, dx - \frac{1}{2} \int_{-\pi/2}^{\pi/2} \cos 2x \, dx. \] The first term evaluates to: \[ \frac{1}{2} \int_{-\pi/2}^{\pi/2} 1 \, dx = \frac{1}{2} \left[ x \right]_{-\pi/2}^{\pi/2} = \frac{1}{2} (\pi - (-\pi)) = \frac{\pi}{2}. \] The second term evaluates to 0 because \( \cos 2x \) is an odd function. Thus: \[ \int_{-\pi/2}^{\pi/2} \sin^2 x \, dx = \frac{\pi}{2}. \]