integral of sin^8(3x)cos^5(3x)

Lösung

\int sin^{8} (3 x) cos^{5} (3 x) d x

\frac{1}{3} (\frac{sin ^{9} ( 3 x )}{9} - \frac{2 sin ^{11} ( 3 x )}{11} + \frac{sin ^{13} ( 3 x )}{13}) + C

Schritte zur Lösung

\int sin^{8} (3 x) cos^{5} (3 x) d x

Wende U-Substitution an

= \int sin^{8} (u) cos^{5} (u) \frac{1}{3} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int sin^{8} (u) cos^{5} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \frac{1}{3} \cdot \int (1 - sin^{2} (u))^{2} cos (u) sin^{8} (u) d u

Wende U-Substitution an

= \frac{1}{3} \cdot \int v^{8} (1 - v^{2})^{2} d v

Multipliziere aus

v^{8} (1 - v^{2})^{2} : v^{8} - 2 v^{10} + v^{12}

= \frac{1}{3} \cdot \int v^{8} - 2 v^{10} + v^{12} d v

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int v^{8} d v - \int 2 v^{10} d v + \int v^{12} d v)

\int v^{8} d v = \frac{v ^{9}}{9}

\int 2 v^{10} d v = \frac{2 v ^{11}}{11}

\int v^{12} d v = \frac{v ^{13}}{13}

= \frac{1}{3} (\frac{v ^{9}}{9} - \frac{2 v ^{11}}{11} + \frac{v ^{13}}{13})

Ersetze zurück

= \frac{1}{3} (\frac{sin ^{9} ( 3 x )}{9} - \frac{2 sin ^{11} ( 3 x )}{11} + \frac{sin ^{13} ( 3 x )}{13})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} (\frac{sin ^{9} ( 3 x )}{9} - \frac{2 sin ^{11} ( 3 x )}{11} + \frac{sin ^{13} ( 3 x )}{13}) + C

d er i v a t i v e f (x) = ln (x^{2} - 8 x + 16)

\frac{\partial}{\partial x} (cos^{2} (θ))

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

\frac{d y}{d x} = 6 x e^{y}

d er i v a t i v e C (x) = - 9 x^{2} + 500 x + 110