integral of t^3(3t^2-4)^{5/2}

Lösung

\int t^{3} (3 t^{2} - 4)^{\frac{5}{2}} d t

\frac{512}{9} (\frac{1}{896} (3 t^{2} - 4)^{\frac{7}{2}} + \frac{1}{4608} (3 t^{2} - 4)^{\frac{9}{2}}) + C

Schritte zur Lösung

\int t^{3} (3 t^{2} - 4)^{\frac{5}{2}} d t

Trigonometrische Substitution anwenden

= \int \frac{512 tan ^{6} ( u ) sec ^{4} ( u )}{9} d u

Entferne die Konstante:

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

= \frac{512}{9} \cdot \int tan^{6} (u) sec^{4} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \frac{512}{9} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{6} (u) d u

Wende U-Substitution an

= \frac{512}{9} \cdot \int v^{6} (1 + v^{2}) d v

Multipliziere aus

v^{6} (1 + v^{2}) : v^{6} + v^{8}

= \frac{512}{9} \cdot \int v^{6} + v^{8} 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{512}{9} (\int v^{6} d v + \int v^{8} d v)

\int v^{6} d v = \frac{v ^{7}}{7}

\int v^{8} d v = \frac{v ^{9}}{9}

= \frac{512}{9} (\frac{v ^{7}}{7} + \frac{v ^{9}}{9})

Ersetze zurück

= \frac{512}{9} \frac{tan ^{7} ( arcsec ( \frac{3 ^{\frac{1}{2}}}{2} t ) )}{7} + \frac{tan ^{9} ( arcsec ( \frac{3 ^{\frac{1}{2}}}{2} t ) )}{9}

Vereinfache

\frac{512}{9} \frac{tan ^{7} ( arcsec ( \frac{3 ^{\frac{1}{2}}}{2} t ) )}{7} + \frac{tan ^{9} ( arcsec ( \frac{3 ^{\frac{1}{2}}}{2} t ) )}{9} : \frac{512}{9} (\frac{1}{896} (3 t^{2} - 4)^{\frac{7}{2}} + \frac{1}{4608} (3 t^{2} - 4)^{\frac{9}{2}})

= \frac{512}{9} (\frac{1}{896} (3 t^{2} - 4)^{\frac{7}{2}} + \frac{1}{4608} (3 t^{2} - 4)^{\frac{9}{2}})

Füge eine Konstante zur Lösung hinzu

= \frac{512}{9} (\frac{1}{896} (3 t^{2} - 4)^{\frac{7}{2}} + \frac{1}{4608} (3 t^{2} - 4)^{\frac{9}{2}}) + C

\int_{6}^{\infty} x e^{- 4 x} d x

e x p an d (3 x^{2} + 3 x - 1)^{4}

\int sin (\frac{x}{y}) d x

\frac{\partial}{\partial x} (x^{8 y})

n = 1 \sum \infty ln^{n} (3)