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

Applying the trigonometric identity \[\sin A \cos B = \frac{1}{2} \left[ \sin(A + B) + \sin(A - B) \right],\] the integral transforms to \[\int_{0}^{\pi/2} \sin 2x \cos 3x \, dx = \frac{1}{2} \int_{0}^{\pi/2} \left[ \sin(5x) + \sin(-x) \right] \, dx.\]Since \[\sin(-x) = -\sin(x),\] the integral becomes \[\frac{1}{2} \left[ \int_{0}^{\pi/2} \sin(5x) \, dx - \int_{0}^{\pi/2} \sin(x) \, dx \right].\]First, evaluate \( \int \sin(5x) \, dx \): \[\int \sin(5x) \, dx = -\frac{1}{5} \cos(5x).\]At the limits, \[\int_{0}^{\pi/2} \sin(5x) \, dx = -\frac{1}{5} \left[ \cos(5 \cdot \pi/2) - \cos(0) \right] = -\frac{1}{5} \left[ 0 - 1 \right] = \frac{1}{5}.\]Next, evaluate \( \int \sin(x) \, dx \): \[\int \sin(x) \, dx = -\cos(x).\]At the limits, \[\int_{0}^{\pi/2} \sin(x) \, dx = -\left[ \cos(\pi/2) - \cos(0) \right] = -\left[ 0 - 1 \right] = 1.\]Substituting these values back, we get \[\int_{0}^{\pi/2} \sin 2x \cos 3x \, dx = \frac{1}{2} \left[ \frac{1}{5} - 1 \right] = \frac{1}{2} \left[ -\frac{4}{5} \right] = -\frac{2}{5}.\] Final Answer: \( \boxed{-\frac{2}{5}} \)