integral of sin^3(6θ)sqrt(cos(6θ))

Lösung

\int sin^{3} (6 θ) cos (6 θ) d θ

- \frac{1}{6} (\frac{2}{3} cos^{\frac{3}{2}} (6 θ) - \frac{2}{7} cos^{\frac{7}{2}} (6 θ)) + C

Schritte zur Lösung

\int sin^{3} (6 θ) cos (6 θ) d θ

Wende U-Substitution an

= \int sin^{3} (u) cos (u) \frac{1}{6} d u

Entferne die Konstante:

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

= \frac{1}{6} \cdot \int sin^{3} (u) cos (u) d u

Wende U-Substitution an

= \frac{1}{6} \cdot \int - v (1 - v^{2}) d v

Entferne die Konstante:

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

= \frac{1}{6} (- \int v (1 - v^{2}) d v)

Multipliziere aus

v (1 - v^{2}) : v - v^{\frac{5}{2}}

= \frac{1}{6} (- \int v - v^{\frac{5}{2}} 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}{6} (- (\int v d v - \int v^{\frac{5}{2}} d v))

\int v d v = \frac{2}{3} v^{\frac{3}{2}}

\int v^{\frac{5}{2}} d v = \frac{2}{7} v^{\frac{7}{2}}

= \frac{1}{6} (- (\frac{2}{3} v^{\frac{3}{2}} - \frac{2}{7} v^{\frac{7}{2}}))

Ersetze zurück

= \frac{1}{6} (- (\frac{2}{3} cos^{\frac{3}{2}} (6 θ) - \frac{2}{7} cos^{\frac{7}{2}} (6 θ)))

Vereinfache

= - \frac{1}{6} (\frac{2}{3} cos^{\frac{3}{2}} (6 θ) - \frac{2}{7} cos^{\frac{7}{2}} (6 θ))

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{6} (\frac{2}{3} cos^{\frac{3}{2}} (6 θ) - \frac{2}{7} cos^{\frac{7}{2}} (6 θ)) + C

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