Question

What is the value of \( \int_0^{\pi/2} \sin x \cos x \, dx\)?

• A. 0
• B. 1/2
• C. 1
• D. 1/4

Hint:

Use the identity \( \sin x \cos x = \frac{1}{2} \sin 2x \) to simplify integrals involving products of sine and cosine.

Answer 1

The Correct Option is B

Step 1: Use the trigonometric identity \( \sin x \cos x = \frac{1}{2} \sin 2x \) to simplify the integrand: \[ \int \sin x \cos x \, dx = \int \frac{1}{2} \sin 2x \, dx. \]
Step 2: Factor out the constant and set up the definite integral: \[ \int_0^{\pi/2} \sin x \cos x \, dx = \frac{1}{2} \int_0^{\pi/2} \sin 2x \, dx. \]
Step 3: Compute the integral. The antiderivative of \( \sin 2x \) is \( -\frac{1}{2} \cos 2x \): \[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C. \]
Step 4: Evaluate the definite integral from 0 to \( \pi/2 \): \[ \frac{1}{2} \left[ -\frac{1}{2} \cos 2x \right]_0^{\pi/2}. \]
Step 5: Substitute the limits: - At \( x = \pi/2 \), \( 2x = \pi \), so \( \cos \pi = -1 \): \[ -\frac{1}{2} \cos(\pi) = -\frac{1}{2} (-1) = \frac{1}{2}. \] - At \( x = 0 \), \( 2x = 0 \), so \( \cos 0 = 1 \): \[ -\frac{1}{2} \cos(0) = -\frac{1}{2} (1) = -\frac{1}{2}. \] - Difference: \[ \frac{1}{2} \left[ \frac{1}{2} - \left(-\frac{1}{2}\right) \right] = \frac{1}{2} \left[ \frac{1}{2} + \frac{1}{2} \right] = \frac{1}{2} \cdot 1 = \frac{1}{2}. \]