integral of sin^3(2x)(1-2cos(2x))^2

解

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

8 (- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6}) - \frac{1}{2} sin^{4} (2 x) + 2 (- \frac{1}{3} cos^{3} (2 x) + \frac{1}{5} cos^{5} (2 x)) + C

解答ステップ

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

拡張

sin^{3} (2 x) (1 - 2 cos (2 x))^{2} : sin^{3} (2 x) - 4 sin^{3} (2 x) cos (2 x) + 4 sin^{3} (2 x) cos^{2} (2 x)

= \int sin^{3} (2 x) - 4 sin^{3} (2 x) cos (2 x) + 4 sin^{3} (2 x) cos^{2} (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 sin^{3} (2 x) d x - \int 4 sin^{3} (2 x) cos (2 x) d x + \int 4 sin^{3} (2 x) cos^{2} (2 x) d x

\int sin^{3} (2 x) d x = 8 (- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6})

\int 4 sin^{3} (2 x) cos (2 x) d x = \frac{1}{2} sin^{4} (2 x)

\int 4 sin^{3} (2 x) cos^{2} (2 x) d x = 2 (- \frac{1}{3} cos^{3} (2 x) + \frac{1}{5} cos^{5} (2 x))

= 8 (- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6}) - \frac{1}{2} sin^{4} (2 x) + 2 (- \frac{1}{3} cos^{3} (2 x) + \frac{1}{5} cos^{5} (2 x))

定数を解答に追加する

= 8 (- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6}) - \frac{1}{2} sin^{4} (2 x) + 2 (- \frac{1}{3} cos^{3} (2 x) + \frac{1}{5} cos^{5} (2 x)) + C