integral of sqrt(cos(3x))sin^3(3x)

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende U-Substitution an

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

Multipliziere aus

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

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

Ersetze zurück

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

\int 8 x + 2 d x

n = 1 \sum \infty n^{7} (1 + n^{8})^{2}

\frac{d}{d x} (\frac{tan ( x )}{1 + tan ( x )})

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

a re a 2 x, x^{2} + 3 x - 6