integral of sin^3(xco)s^5x

Lösung

\int sin^{3} (x co) s^{5} x d x

s^{5} \frac{1}{c ^{2} o ^{2}} (- co x cos (co x) + \frac{1}{3} co x cos^{3} (co x) + \frac{2}{3} sin (co x) + \frac{1}{9} sin^{3} (co x)) + C

Schritte zur Lösung

\int sin^{3} (x co) s^{5} x d x

Entferne die Konstante:

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

= s^{5} \cdot \int sin^{3} (co x) x d x

Wende U-Substitution an

= s^{5} \cdot \int \frac{u sin ^{3} ( u )}{c ^{2} o ^{2}} d u

Entferne die Konstante:

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

= s^{5} \frac{1}{c ^{2} o ^{2}} \cdot \int u sin^{3} (u) d u

Wende die partielle Integration an

= s^{5} \frac{1}{c ^{2} o ^{2}} (u (- cos (u) + \frac{1}{3} cos^{3} (u)) - \int - cos (u) + \frac{1}{3} cos^{3} (u) d u)

\int - cos (u) + \frac{1}{3} cos^{3} (u) d u = - \frac{2}{3} sin (u) - \frac{1}{9} sin^{3} (u)

= s^{5} \frac{1}{c ^{2} o ^{2}} (u (- cos (u) + \frac{1}{3} cos^{3} (u)) - (- \frac{2}{3} sin (u) - \frac{1}{9} sin^{3} (u)))

Setze in

u = co x

ein

= s^{5} \frac{1}{c ^{2} o ^{2}} (co x (- cos (co x) + \frac{1}{3} cos^{3} (co x)) - (- \frac{2}{3} sin (co x) - \frac{1}{9} sin^{3} (co x)))

Multipliziere aus

co x (- cos (co x) + \frac{1}{3} cos^{3} (co x)) - (- \frac{2}{3} sin (co x) - \frac{1}{9} sin^{3} (co x)) : - co x cos (co x) + \frac{1}{3} co x cos^{3} (co x) + \frac{2}{3} sin (co x) + \frac{1}{9} sin^{3} (co x)

= s^{5} \frac{1}{c ^{2} o ^{2}} (- co x cos (co x) + \frac{1}{3} co x cos^{3} (co x) + \frac{2}{3} sin (co x) + \frac{1}{9} sin^{3} (co x))

Füge eine Konstante zur Lösung hinzu

= s^{5} \frac{1}{c ^{2} o ^{2}} (- co x cos (co x) + \frac{1}{3} co x cos^{3} (co x) + \frac{2}{3} sin (co x) + \frac{1}{9} sin^{3} (co x)) + C

d er i v a t i v e f (x) = tan (2 x) \cdot sin (2 x)

\int_{- 3}^{3} (18 - 2 x^{2}) d x

\int x^{3} e^{x^{4}} d x

6 y^{^{''}} + 12 y^{^{'}} - 18 y = 0

\frac{\partial}{\partial x} ((x - 2) (x^{2} + 4))