ปริพันธ์ ฟังก์ชันตรีโกณมิติ

$$ \int \sin ^2 x \cos ^4 x dx \\ $$
วิธีทำ \begin{align} & \cssId{Step1}{ \sin ^2 x \cos ^4 x = (\frac{1-\cos 2x }{2} ) (\frac{1+\cos 2x}{2})^2 }\\ & \cssId{Step2} {= ( \frac{1-\cos 2x }{2})( \frac{1+2 \cos 2x + \cos ^2 2x}{4} ) }\\ & \cssId{Step3} { = \frac{1+\cos 2x - \cos ^2 2x - \cos ^3 2x}{8} } \\ & \cssId{Step4}{ = \frac{1+\cos 2x - \frac{1+\cos 4x}{2} - \cos ^3 2x}{8} } \\ & \cssId{Step5}{ = \frac{1+\cos 2x - \frac{1+\cos 4x}{2} - (1- \sin^2 2x) \cos 2x}{8} } \\ &\cssId{Step6}{ ดังนั้น \int \sin ^2 x \cos ^4 x dx = \frac{1}{8} \int [ {1+\cos 2x - ( \frac{1+\cos 4x}{2}) - (1- \sin^2 2x) \cos 2x} ] dx } \\ &\cssId{Step7}{= \frac{1}{8} [ x+\frac{\sin 2x}{2} -\frac{ (x+\frac{\sin 4x}{4}) } {2} -(\frac{ \sin 2x}{2} -\frac{\sin^3 2x}{6} )] + C} \\ \end{align}