integral of x^3sqrt(a^2x^2-b^2)

Lösung

\int x^{3} a^{2} x^{2} - b^{2} d x

\frac{5 b ^{2} ( a ^{2} x ^{2} - b ^{2} ) ^{\frac{3}{2}} + 3 ( a ^{2} x ^{2} - b ^{2} ) ^{\frac{5}{2}}}{15 a ^{4}} + C

Schritte zur Lösung

\int x^{3} a^{2} x^{2} - b^{2} d x

Trigonometrische Substitution anwenden

= \int \frac{b ^{5} sec ^{4} ( u ) tan ^{2} ( u )}{a ^{4}} d u

Entferne die Konstante:

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

= \frac{b ^{5}}{a ^{4}} \cdot \int sec^{4} (u) tan^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \frac{b ^{5}}{a ^{4}} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{2} (u) d u

Wende U-Substitution an

= \frac{b ^{5}}{a ^{4}} \cdot \int v^{2} (1 + v^{2}) d v

Multipliziere aus

v^{2} (1 + v^{2}) : v^{2} + v^{4}

= \frac{b ^{5}}{a ^{4}} \cdot \int v^{2} + v^{4} 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{b ^{5}}{a ^{4}} (\int v^{2} d v + \int v^{4} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{b ^{5}}{a ^{4}} (\frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Ersetze zurück

= \frac{b ^{5}}{a ^{4}} (\frac{tan ^{3} ( arcsec ( \frac{a}{b} x ) )}{3} + \frac{tan ^{5} ( arcsec ( \frac{a}{b} x ) )}{5})

Vereinfache

\frac{b ^{5}}{a ^{4}} (\frac{tan ^{3} ( arcsec ( \frac{a}{b} x ) )}{3} + \frac{tan ^{5} ( arcsec ( \frac{a}{b} x ) )}{5}) : \frac{5 b ^{2} ( a ^{2} x ^{2} - b ^{2} ) ^{\frac{3}{2}} + 3 ( a ^{2} x ^{2} - b ^{2} ) ^{\frac{5}{2}}}{15 a ^{4}}

= \frac{5 b ^{2} ( a ^{2} x ^{2} - b ^{2} ) ^{\frac{3}{2}} + 3 ( a ^{2} x ^{2} - b ^{2} ) ^{\frac{5}{2}}}{15 a ^{4}}

Füge eine Konstante zur Lösung hinzu

= \frac{5 b ^{2} ( a ^{2} x ^{2} - b ^{2} ) ^{\frac{3}{2}} + 3 ( a ^{2} x ^{2} - b ^{2} ) ^{\frac{5}{2}}}{15 a ^{4}} + C

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