Question

Find the value of the integral: \[ \int_0^\pi \sin^2(x) \, dx. \]

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

Hint:

Use trigonometric identities to simplify integrals involving trigonometric functions, and always check the symmetry of the integrand.

Answer 1

The Correct Option is B

We will use the identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \) to simplify the integral: \[ \int_0^\pi \sin^2(x) \, dx = \int_0^\pi \frac{1 - \cos(2x)}{2} \, dx. \] Now, split the integral: \[ \int_0^\pi \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \int_0^\pi 1 \, dx - \frac{1}{2} \int_0^\pi \cos(2x) \, dx. \] The first integral is: \[ \int_0^\pi 1 \, dx = \pi. \] The second integral is: \[ \int_0^\pi \cos(2x) \, dx = 0 \quad \text{(since \( \cos(2x) \) is symmetric about \( \pi/2 \))}. \] Thus, the integral becomes: \[ \frac{1}{2} \times \pi = \frac{\pi}{2}. \] The correct answer is: \[ \boxed{\frac{\pi}{2}}. \]