integral of sin^3(4θ)sqrt(cos(4θ))

Lösung

\int sin^{3} (4 θ) cos (4 θ) d θ

- \frac{1}{4} (\frac{2}{3} cos^{\frac{3}{2}} (4 θ) - \frac{2}{7} cos^{\frac{7}{2}} (4 θ)) + C

Schritte zur Lösung

\int sin^{3} (4 θ) cos (4 θ) d θ

Wende U-Substitution an

= \int sin^{3} (u) cos (u) \frac{1}{4} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int sin^{3} (u) cos (u) d u

Wende U-Substitution an

= \frac{1}{4} \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}{4} (- \int v (1 - v^{2}) d v)

Multipliziere aus

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

= \frac{1}{4} (- \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}{4} (- (\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}{4} (- (\frac{2}{3} v^{\frac{3}{2}} - \frac{2}{7} v^{\frac{7}{2}}))

Ersetze zurück

= \frac{1}{4} (- (\frac{2}{3} cos^{\frac{3}{2}} (4 θ) - \frac{2}{7} cos^{\frac{7}{2}} (4 θ)))

Vereinfache

= - \frac{1}{4} (\frac{2}{3} cos^{\frac{3}{2}} (4 θ) - \frac{2}{7} cos^{\frac{7}{2}} (4 θ))

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{4} (\frac{2}{3} cos^{\frac{3}{2}} (4 θ) - \frac{2}{7} cos^{\frac{7}{2}} (4 θ)) + C

t an g e n t f (x) = x^{2} + 4

\frac{d y}{d x} = \frac{1 + x y}{1 + x}

\frac{\partial}{\partial y} (x sin (x y))

\frac{d}{d x} (sin (2 x - 1))

\int - a cos (x) d x