integral of (sin^3(2t))sqrt(cos(2t))

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende U-Substitution an

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

Multipliziere aus

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

(3 x^{2})^{^{'}}

\int \frac{x ^{2}}{x ^{6} + 1} d x

d er i v a t i v e f (x) = ln (x + x^{2} + 1)

\int (7 + e) d x

\int \frac{( x ^{2} )}{x ^{2} + 64} d x