Question

Evaluate the definite integral: \[ \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx \]

• A. \( \dfrac{\pi}{2} \)
• B. \( \dfrac{\pi}{4} \)
• C. \( \dfrac{\pi}{8} \)
• D. \( \pi \)

Hint:

Useful definite integrals: \[ \int_{0}^{\frac{\pi}{2}}\sin^2 x\,dx = \int_{0}^{\frac{\pi}{2}}\cos^2 x\,dx = \frac{\pi}{4} \]

Answer 1

The Correct Option is B

Concept: Use the trigonometric identity \[ \sin^2 x = \frac{1-\cos 2x}{2} \] This identity simplifies the integration.
Step 1: {Substitute the identity into the integral.} \[ \int_{0}^{\frac{\pi}{2}} \sin^2 x\,dx = \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2x}{2}\,dx \]
Step 2: {Split the integral.} \[ = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}1\,dx - \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\cos 2x\,dx \]
Step 3: {Evaluate both integrals.} \[ \frac{1}{2}\int_{0}^{\frac{\pi}{2}}1\,dx = \frac{1}{2}\left[\frac{\pi}{2}\right] = \frac{\pi}{4} \] \[ \int_{0}^{\frac{\pi}{2}}\cos 2x\,dx = \left[\frac{\sin 2x}{2}\right]_{0}^{\frac{\pi}{2}}=0 \]
Step 4: {Final result.} \[ \int_{0}^{\frac{\pi}{2}} \sin^2 x\,dx = \frac{\pi}{4} \] Thus, the correct option is (B).