Question

What is \(\int \sin 2x\cos 3x\,dx\)?

Options

• \(-\frac{1}{5}\cos 5x+\cos x+\text{C}\)

• \(\frac{1}{10}\cos 5x-\frac{1}{2}\cos x+\text{C}\)

• \(-\frac{1}{10}\cos 5x+\frac{1}{2}\cos x+\text{C}\)

• \(-\frac{1}{10}\sin 5x+\frac{1}{2}\sin x+\text{C}\)

Answer 1

First rewrite the product as \(\frac{1}{2}[\sin 5x-\sin x]\). Integrating term by term gives \(\frac{1}{2}[-\frac{1}{5}\cos 5x+\cos x]+\text{C}\), which simplifies to the stated result.