Question

\(\int_0^{\frac{\pi}{2}} \sin x \cos x \, dx \)

• A. \( 1 \)
• B. \( \frac{1}{2} \)
• C. \( -1 \)
• D. \( \frac{1}{4} \)

Hint:

Use the identity \( \sin(2x) = 2 \sin x \cos x \) to simplify integrals involving \( \sin x \cos x \).

Answer 1

The Correct Option is B

We are given: \[ I = \int_0^{\frac{\pi}{2}} \sin x \cos x \, dx \] Using the trigonometric identity \( \sin(2x) = 2 \sin x \cos x \), we can rewrite the integral as: \[ I = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(2x) \, dx \] Now, the integral of \( \sin(2x) \) is \( -\frac{1}{2} \cos(2x) \), so: \[ I = \frac{1}{2} \left[ -\frac{1}{2} \cos(2x) \right]_0^{\frac{\pi}{2}} = \frac{1}{2} \left( -\frac{1}{2} (\cos(\pi) - \cos(0)) \right) \] Since \( \cos(\pi) = -1 \) and \( \cos(0) = 1 \), we have: \[ I = \frac{1}{2} \left( -\frac{1}{2} (-1 - 1) \right) = \frac{1}{2} \left( \frac{1}{2} \times 2 \right) = \frac{1}{2} \] Thus, the correct answer is \( \frac{1}{2} \).