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

Lösung

\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

Schritte zur Lösung

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

Multipliziere aus

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

Wende die Summenregel an:

\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)

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

x \to 2 lim (\frac{1}{x ^{2} - 4 x + 4})

in v erse l a pl a ce \frac{5}{s} - \frac{5}{s} e^{- 5 s}

n = 0 \sum \infty - (- 1)^{n + 1}

\frac{d}{d x} (\frac{3}{4 x ^{2}})

t an g e n t f (x) = x^{4} - 13 x^{2} + 36