Question

Evaluate: \(\displaystyle \int_0^{\frac{\pi}{2}} \cos 2x \, dx = \)

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(2\)

Hint:

Remember: \(\int \cos(kx) \, dx = \frac{\sin(kx)}{k} + c\). For definite integrals, always carefully substitute the limits.

Answer 1

The Correct Option is B

Step 1: Integrate \(\cos 2x\): \[ \int \cos 2x \, dx = \frac{\sin 2x}{2} + c \] Step 2: Evaluate definite integral: \[ \int_0^{\frac{\pi}{2}} \cos 2x \, dx = \left[ \frac{\sin 2x}{2} \right]_0^{\frac{\pi}{2}} = \frac{\sin \pi}{2} - \frac{\sin 0}{2} = 0 - 0 = 0 \] Wait, result is 0, which contradicts option (B). Let's carefully check: \[ \sin \pi = 0, \quad \sin 0 = 0 \implies \text{integral} = 0 \] So correct answer is (A) 0. --- Correction: Correct Answer: (A) \(0\)