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We use the trigonometric identity: $sin~A~cos~B = \frac{1}{2}[sin(A+B) + sin(A-B)]$. Applying this to the integrand, we have: $sin~2x~cos~3x = \frac{1}{2}[sin(2x+3x) + sin(2x-3x)] = \frac{1}{2}[sin~5x + sin(-x)] = \frac{1}{2}[sin~5x - sin~x]$
Now, we rewrite the integral using the identity from Step 1: $\int_{0}^{\pi/2}sin~2x~cos~3x~dx = \int_{0}^{\pi/2} \frac{1}{2}[sin~5x - sin~x]~dx = \frac{1}{2}\int_{0}^{\pi/2} (sin~5x - sin~x)~dx$
We integrate term by term: $\frac{1}{2}\int_{0}^{\pi/2} (sin~5x - sin~x)~dx = \frac{1}{2} \left[ \int_{0}^{\pi/2} sin~5x~dx - \int_{0}^{\pi/2} sin~x~dx \right]$ $= \frac{1}{2} \left[ -\frac{cos~5x}{5} \Big|_{0}^{\pi/2} + cos~x \Big|_{0}^{\pi/2} \right]$
Now, we evaluate the limits: $= \frac{1}{2} \left[ -\frac{cos(5\pi/2)}{5} + \frac{cos(0)}{5} + cos(\pi/2) - cos(0) \right]$ Since $cos(5\pi/2) = cos(\pi/2) = 0$, $cos(0) = 1$, we have: $= \frac{1}{2} \left[ -\frac{0}{5} + \frac{1}{5} + 0 - 1 \right] = \frac{1}{2} \left[ \frac{1}{5} - 1 \right] = \frac{1}{2} \left[ \frac{1-5}{5} \right] = \frac{1}{2} \left[ \frac{-4}{5} \right] = -\frac{2}{5}$
Final Answer: -2/5