The definite integral \( \int_0^{\pi} \sin^2 x \, dx \) is to be evaluated. Step 1: Apply a standard trigonometric identity Utilize the identity for \( \sin^2 x \): \[ \sin^2 x = \frac{1 - \cos(2x)}{2} \] The integral transforms to: \[ \int_0^{\pi} \sin^2 x \, dx = \int_0^{\pi} \frac{1 - \cos(2x)}{2} \, dx \] Step 2: Decompose the integral \[ = \frac{1}{2} \int_0^{\pi} 1 \, dx - \frac{1}{2} \int_0^{\pi} \cos(2x) \, dx \] Step 3: Compute the integrals - The first integral is: \[ \int_0^{\pi} 1 \, dx = x \Big|_0^{\pi} = \pi \] - The second integral is: \[ \int_0^{\pi} \cos(2x) \, dx = \frac{\sin(2x)}{2} \Big|_0^{\pi} = \frac{\sin(2\pi)}{2} - \frac{\sin(0)}{2} = 0 \] Step 4: Consolidate the results Substitute the computed values back: \[ \int_0^{\pi} \sin^2 x \, dx = \frac{1}{2} \times \pi - \frac{1}{2} \times 0 = \frac{\pi}{2} \] The integral evaluates to \( \frac{\pi}{2} \). Answer: The integral's value is \( \frac{\pi}{2} \), corresponding to option (1).