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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

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

定数を解答に追加する

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