integral of 19sin^3(xco)s^2x

Lösung

\int 19 sin^{3} (x co) s^{2} x d x

s^{2} \frac{19}{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 19 sin^{3} (x co) s^{2} x d x

Entferne die Konstante:

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

= 19 s^{2} \cdot \int sin^{3} (co x) x d x

Wende U-Substitution an

= 19 s^{2} \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

= 19 s^{2} \frac{1}{c ^{2} o ^{2}} \cdot \int u sin^{3} (u) d u

Wende die partielle Integration an

= 19 s^{2} \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)

= 19 s^{2} \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

= 19 s^{2} \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)))

Vereinfache

19 s^{2} \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))) : s^{2} \frac{19}{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))

= s^{2} \frac{19}{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^{2} \frac{19}{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

t an g e n t y = x^{2} + x, (- 1, 0)

\frac{d}{d x} (\frac{x ^{2}}{( x ^{2} - 1 ) ^{4}})

f (x) = \frac{t ^{3}}{6} - \frac{2}{t ^{2}}

t^{2} y^{^{''}} + 19 t y^{^{'}} + 90 y = 0

- 2 y^{^{''}} + 8 y = 0