integral of sin^{1/2}(2z)cos^3(2z)

Lösung

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

\frac{1}{3} sin^{\frac{3}{2}} (2 z) - \frac{1}{7} sin^{\frac{7}{2}} (2 z) + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Wende U-Substitution an

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

Multipliziere aus

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

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

Ersetze zurück

= \frac{1}{2} (\frac{2}{3} sin^{\frac{3}{2}} (2 z) - \frac{2}{7} sin^{\frac{7}{2}} (2 z))

Vereinfache

\frac{1}{2} (\frac{2}{3} sin^{\frac{3}{2}} (2 z) - \frac{2}{7} sin^{\frac{7}{2}} (2 z)) : \frac{1}{3} sin^{\frac{3}{2}} (2 z) - \frac{1}{7} sin^{\frac{7}{2}} (2 z)

= \frac{1}{3} sin^{\frac{3}{2}} (2 z) - \frac{1}{7} sin^{\frac{7}{2}} (2 z)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} sin^{\frac{3}{2}} (2 z) - \frac{1}{7} sin^{\frac{7}{2}} (2 z) + C

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