integral of sin^3(2θ)cos^{3/2}(2θ)

Lösung

\int sin^{3} (2 θ) cos^{\frac{3}{2}} (2 θ) d θ

- \frac{1}{5} cos^{\frac{5}{2}} (2 θ) + \frac{1}{9} cos^{\frac{9}{2}} (2 θ) + C

Schritte zur Lösung

\int sin^{3} (2 θ) cos^{\frac{3}{2}} (2 θ) d θ

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Wende U-Substitution an

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

Multipliziere aus

- v^{\frac{3}{2}} (1 - v^{2}) : - v^{\frac{3}{2}} + v^{\frac{7}{2}}

= \frac{1}{2} \cdot \int - v^{\frac{3}{2}} + v^{\frac{7}{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}{2} (- \int v^{\frac{3}{2}} d v + \int v^{\frac{7}{2}} d v)

\int v^{\frac{3}{2}} d v = \frac{2}{5} v^{\frac{5}{2}}

\int v^{\frac{7}{2}} d v = \frac{2}{9} v^{\frac{9}{2}}

= \frac{1}{2} (- \frac{2}{5} v^{\frac{5}{2}} + \frac{2}{9} v^{\frac{9}{2}})

Ersetze zurück

= \frac{1}{2} (- \frac{2}{5} cos^{\frac{5}{2}} (2 θ) + \frac{2}{9} cos^{\frac{9}{2}} (2 θ))

Vereinfache

\frac{1}{2} (- \frac{2}{5} cos^{\frac{5}{2}} (2 θ) + \frac{2}{9} cos^{\frac{9}{2}} (2 θ)) : - \frac{1}{5} cos^{\frac{5}{2}} (2 θ) + \frac{1}{9} cos^{\frac{9}{2}} (2 θ)

= - \frac{1}{5} cos^{\frac{5}{2}} (2 θ) + \frac{1}{9} cos^{\frac{9}{2}} (2 θ)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{5} cos^{\frac{5}{2}} (2 θ) + \frac{1}{9} cos^{\frac{9}{2}} (2 θ) + C

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