Question

The integral $ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx \text{ is equal to} $

• A. 0
• B. \( \pi \)
• C. \( \frac{\pi}{2} \)
• D. \( \frac{\pi}{4} \)

Hint:

For integrals involving \( \sin^2 x \), you can use the identity \( \sin^2 x = \frac{1}{2} (1 - \cos(2x)) \) to simplify the expression before integrating.

Answer 1

The Correct Option is C

To evaluate the integral: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx \] We can use the identity \( \sin^2 x = \frac{1}{2} (1 - \cos(2x)) \) to simplify the integrand: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2} (1 - \cos(2x)) \, dx \] This separates into two integrals: \[ I = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx - \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(2x) \, dx \] The first integral evaluates to: \[ \frac{1}{2} \left( x \bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \right) = \frac{1}{2} \left( \frac{\pi}{2} - \left( -\frac{\pi}{2} \right) \right) = \frac{\pi}{2} \] The second integral is: \[ \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(2x) \, dx \] The integral of \( \cos(2x) \) is \( \frac{1}{2} \sin(2x) \), and since \( \sin(2x) \) is zero at both bounds \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \), this part evaluates to zero: \[ \frac{1}{2} \left[ \frac{1}{2} \sin(2x) \bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \right] = 0 \] Thus, the total value of the integral is: \[ I = \frac{\pi}{2} \]