integral of cos(x)(2+sin(x))^6

Lösung

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

64 sin (x) + 96 sin^{2} (x) + 80 sin^{3} (x) + 40 sin^{4} (x) + 12 sin^{5} (x) + 2 sin^{6} (x) + \frac{sin ^{7} ( x )}{7} + C

Schritte zur Lösung

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

Multipliziere aus

cos (x) (2 + sin (x))^{6} : 64 cos (x) + 192 cos (x) sin (x) + 240 sin^{2} (x) cos (x) + 160 sin^{3} (x) cos (x) + 60 sin^{4} (x) cos (x) + 12 sin^{5} (x) cos (x) + sin^{6} (x) cos (x)

= \int 64 cos (x) + 192 cos (x) sin (x) + 240 sin^{2} (x) cos (x) + 160 sin^{3} (x) cos (x) + 60 sin^{4} (x) cos (x) + 12 sin^{5} (x) cos (x) + sin^{6} (x) cos (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 64 cos (x) d x + \int 192 cos (x) sin (x) d x + \int 240 sin^{2} (x) cos (x) d x + \int 160 sin^{3} (x) cos (x) d x + \int 60 sin^{4} (x) cos (x) d x + \int 12 sin^{5} (x) cos (x) d x + \int sin^{6} (x) cos (x) d x

\int 64 cos (x) d x = 64 sin (x)

\int 192 cos (x) sin (x) d x = 96 sin^{2} (x)

\int 240 sin^{2} (x) cos (x) d x = 80 sin^{3} (x)

\int 160 sin^{3} (x) cos (x) d x = 40 sin^{4} (x)

\int 60 sin^{4} (x) cos (x) d x = 12 sin^{5} (x)

\int 12 sin^{5} (x) cos (x) d x = 2 sin^{6} (x)

\int sin^{6} (x) cos (x) d x = \frac{sin ^{7} ( x )}{7}

= 64 sin (x) + 96 sin^{2} (x) + 80 sin^{3} (x) + 40 sin^{4} (x) + 12 sin^{5} (x) + 2 sin^{6} (x) + \frac{sin ^{7} ( x )}{7}

Füge eine Konstante zur Lösung hinzu

= 64 sin (x) + 96 sin^{2} (x) + 80 sin^{3} (x) + 40 sin^{4} (x) + 12 sin^{5} (x) + 2 sin^{6} (x) + \frac{sin ^{7} ( x )}{7} + C

\frac{d y}{d t} + t y^{2} = 0

d er i v a t i v e \frac{4 x}{e ^{x}}

\frac{d}{d x} (\frac{1 + 1}{x})

a re a 1 - x^{2}, ∣ x ∣

\frac{d}{d x} (4 1 + x)