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

Lösung

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

\frac{3}{8} x - \frac{3}{32} sin (4 x) + \frac{19}{24} cos^{3} (x) sin (x) - \frac{1}{6} cos^{5} (x) sin (x) - \frac{5}{32} (2 x + sin (2 x)) + C

Schritte zur Lösung

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

Vereinfache

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Multipliziere aus

cos^{4} (x) (1 - cos^{2} (x)) : cos^{4} (x) - cos^{6} (x)

= \int cos^{4} (x) - cos^{6} (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^{4} (x) d x - \int cos^{6} (x) d x

\int cos^{4} (x) d x = cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x))

\int cos^{6} (x) d x = \frac{sin ( x ) cos ^{5} ( x )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (x) sin (x) + \frac{3}{8} (x + \frac{1}{2} sin (2 x)))

= cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x)) - (\frac{sin ( x ) cos ^{5} ( x )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (x) sin (x) + \frac{3}{8} (x + \frac{1}{2} sin (2 x))))

Vereinfache

cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x)) - (\frac{sin ( x ) cos ^{5} ( x )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (x) sin (x) + \frac{3}{8} (x + \frac{1}{2} sin (2 x)))) : \frac{3}{8} x - \frac{3}{32} sin (4 x) + \frac{19}{24} cos^{3} (x) sin (x) - \frac{1}{6} cos^{5} (x) sin (x) - \frac{5}{32} (2 x + sin (2 x))

= \frac{3}{8} x - \frac{3}{32} sin (4 x) + \frac{19}{24} cos^{3} (x) sin (x) - \frac{1}{6} cos^{5} (x) sin (x) - \frac{5}{32} (2 x + sin (2 x))

Füge eine Konstante zur Lösung hinzu

= \frac{3}{8} x - \frac{3}{32} sin (4 x) + \frac{19}{24} cos^{3} (x) sin (x) - \frac{1}{6} cos^{5} (x) sin (x) - \frac{5}{32} (2 x + sin (2 x)) + C

- y^{^{'}} - y = e^{x}

n = 1 \sum \infty - 1

\frac{d y}{d x} = y + 2

f (x) = 4 sin (2 x) - 3 cos (3 x)

\int_{\frac{1}{5}}^{7} 14 x ln (5 x) d x