ปริพันธ์ ฟังก์ชันตรีโกณมิติ

$$ \int \sin 2x \sin 3x \cos 4x \quad dx \\ $$
วิธีทำ \begin{align} & \cssId{Step1}{ \sin 2x \sin 3x \cos 4x = \sin 2x ( \frac{\sin 7x - \sin x }{2})}\\ & \cssId{Step2}{ = \frac{1}{2} (\sin 2x \sin 7x - \sin x \sin 2x ) } \\ & \cssId{Step3}{ = \frac{1}{4} (\cos 5x - \cos 9x - ( \cos x - \cos 3x ) ) } \\ & \cssId{Step4}{ = \frac{1}{4} (\cos 5x - \cos 9x - \cos x + \cos 3x ) } \\ & \cssId{Step5}{ดังนั้น \quad \int \sin 2x \sin 3x \cos 4x \quad dx = \int \frac{1}{4} (\cos 5x - \cos 9x - \cos x + \cos 3x ) dx } \\ & \cssId{Step6}{ = \frac{1}{4} (\frac{ \sin 5x}{5} - \frac{\sin 9x}{9} - \sin x + \frac{\sin 3x}{3} ) +C } \\ \end{align}