integral of cos(2x)(1+sin(2x))^4

Lösung

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

\frac{1}{2} sin (2 x) - \frac{1}{2} cos (4 x) + 3 (- \frac{1}{2} sin (2 x) cos (4 x) + \frac{1}{6} sin (6 x)) + \frac{1}{2} sin^{4} (2 x) + \frac{1}{10} sin^{5} (2 x) + C

Schritte zur Lösung

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

Multipliziere aus

cos (2 x) (1 + sin (2 x))^{4} : cos (2 x) + 4 cos (2 x) sin (2 x) + 6 sin^{2} (2 x) cos (2 x) + 4 sin^{3} (2 x) cos (2 x) + sin^{4} (2 x) cos (2 x)

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

Wende die Summenregel an:

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

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

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

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

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

\int sin^{4} (2 x) cos (2 x) d x = \frac{1}{10} sin^{5} (2 x)

= \frac{1}{2} sin (2 x) - \frac{1}{2} cos (4 x) + 3 (- \frac{1}{2} sin (2 x) cos (4 x) + \frac{1}{6} sin (6 x)) + \frac{1}{2} sin^{4} (2 x) + \frac{1}{10} sin^{5} (2 x)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} sin (2 x) - \frac{1}{2} cos (4 x) + 3 (- \frac{1}{2} sin (2 x) cos (4 x) + \frac{1}{6} sin (6 x)) + \frac{1}{2} sin^{4} (2 x) + \frac{1}{10} sin^{5} (2 x) + C

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