Question

Evaluate: \[ \int_{0}^{\pi/2} \sin 2x \cos 3x \, dx \]

Hint:

Use trigonometric identities to simplify products of sine and cosine into sums or differences.

Answer 1

Using the trigonometric identity: \[ \sin A \cos B = \frac{1}{2} \left[ \sin(A + B) + \sin(A - B) \right], \] we rewrite the integral as: \[ \int_{0}^{\pi/2} \sin 2x \cos 3x \, dx = \frac{1}{2} \int_{0}^{\pi/2} \left[ \sin(5x) + \sin(-x) \right] \, dx. \] Simplify: \[ \sin(-x) = -\sin(x), \quad {so the integral becomes:} \] \[ \frac{1}{2} \left[ \int_{0}^{\pi/2} \sin(5x) \, dx - \int_{0}^{\pi/2} \sin(x) \, dx \right]. \] 1. Evaluate \( \int \sin(5x) \, dx \): \[ \int \sin(5x) \, dx = -\frac{1}{5} \cos(5x). \] At the limits: \[ \int_{0}^{\pi/2} \sin(5x) \, dx = -\frac{1}{5} \left[ \cos(5 \cdot \pi/2) - \cos(0) \right] = -\frac{1}{5} \left[ 0 - 1 \right] = \frac{1}{5}. \] 2. Evaluate \( \int \sin(x) \, dx \): \[ \int \sin(x) \, dx = -\cos(x). \] At the limits: \[ \int_{0}^{\pi/2} \sin(x) \, dx = -\left[ \cos(\pi/2) - \cos(0) \right] = -\left[ 0 - 1 \right] = 1. \] Substitute back: \[ \int_{0}^{\pi/2} \sin 2x \cos 3x \, dx = \frac{1}{2} \left[ \frac{1}{5} - 1 \right] = \frac{1}{2} \left[ -\frac{4}{5} \right] = -\frac{2}{5}. \] Final Answer: \( \boxed{-\frac{2}{5}} \)