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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

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

簡素化

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