integral of sin(2x)(3-cos(2x))^2

解

\int sin (2 x) (3 - cos (2 x))^{2} d x

- \frac{9}{2} cos (2 x) + \frac{3}{4} cos (4 x) - \frac{1}{6} cos^{3} (2 x) + C

解答ステップ

\int sin (2 x) (3 - cos (2 x))^{2} d x

拡張

sin (2 x) (3 - cos (2 x))^{2} : 9 sin (2 x) - 6 sin (2 x) cos (2 x) + cos^{2} (2 x) sin (2 x)

= \int 9 sin (2 x) - 6 sin (2 x) cos (2 x) + cos^{2} (2 x) sin (2 x) d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 9 sin (2 x) d x - \int 6 sin (2 x) cos (2 x) d x + \int cos^{2} (2 x) sin (2 x) d x

\int 9 sin (2 x) d x = - \frac{9}{2} cos (2 x)

\int 6 sin (2 x) cos (2 x) d x = - \frac{3}{4} cos (4 x)

\int cos^{2} (2 x) sin (2 x) d x = - \frac{1}{6} cos^{3} (2 x)

= - \frac{9}{2} cos (2 x) - (- \frac{3}{4} cos (4 x)) - \frac{1}{6} cos^{3} (2 x)

簡素化

= - \frac{9}{2} cos (2 x) + \frac{3}{4} cos (4 x) - \frac{1}{6} cos^{3} (2 x)

定数を解答に追加する

= - \frac{9}{2} cos (2 x) + \frac{3}{4} cos (4 x) - \frac{1}{6} cos^{3} (2 x) + C