Question

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

Hint:

Use trigonometric identities to simplify integrals involving products of sine and cosine.

Answer 1

Step 1: {Use the product-to-sum formula}
The product-to-sum formula is: \[ \sin A \cos B = \frac{1}{2}[\sin(A + B) + \sin(A - B)]. \] Substituting \( A = 2x \) and \( B = 3x \): \[ \sin 2x \cos 3x = \frac{1}{2}[\sin(5x) + \sin(-x)] = \frac{1}{2}[\sin(5x) - \sin(x)]. \] Step 2: {Split the integral}
\[ \int_{0}^{\pi/2} \sin 2x \cos 3x \, dx = \frac{1}{2} \int_{0}^{\pi/2} \sin(5x) \, dx - \frac{1}{2} \int_{0}^{\pi/2} \sin(x) \, dx. \] Step 3: {Evaluate the integrals}
\[ \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx). \] For \( \int \sin(5x) \, dx \): \[ \int_{0}^{\pi/2} \sin(5x) \, dx = \left[-\frac{1}{5} \cos(5x)\right]_{0}^{\pi/2} = -\frac{1}{5}[\cos(5\cdot\frac{\pi}{2}) - \cos(0)] = -\frac{1}{5}[0 - 1] = \frac{1}{5}. \] For \( \int \sin(x) \, dx \): \[ \int_{0}^{\pi/2} \sin(x) \, dx = \left[-\cos(x)\right]_{0}^{\pi/2} = -[\cos(\frac{\pi}{2}) - \cos(0)] = -[0 - 1] = 1. \] Step 4: {Combine the results}
\[ \int_{0}^{\pi/2} \sin 2x \cos 3x \, dx = \frac{1}{2} \left(\frac{1}{5} - 1\right) = \frac{1}{2} \cdot -\frac{4}{5} = -\frac{2}{5}. \] Step 5: {Conclude the result}
\[ \int_{0}^{\pi/2} \sin 2x \cos 3x \, dx = -\frac{2}{5}. \]