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

Lösung

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

- \frac{a ^{2} x x ^{2} - a ^{2}}{8} - \frac{a ^{4}}{8} ln (\frac{x ^{2} - a ^{2} + x}{∣ a ∣}) + \frac{1}{4} x^{3} x^{2} - a^{2} + C

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int a^{4} sec^{3} (u) tan^{2} (u) d u

Entferne die Konstante:

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

= a^{4} \cdot \int sec^{3} (u) tan^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= a^{4} \cdot \int sec^{3} (u) (- 1 + sec^{2} (u)) d u

Multipliziere aus

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= a^{4} \cdot \int - sec^{3} (u) + sec^{5} (u) d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= a^{4} (- \int sec^{3} (u) d u + \int sec^{5} (u) d u)

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

\int sec^{5} (u) d u = \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

= a^{4} (- (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣) + \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣))

Setze in

u = arcsec (\frac{1}{a} x)

ein

= a^{4} (- (\frac{1}{2} sec (arcsec (\frac{1}{a} x)) tan (arcsec (\frac{1}{a} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{a} x)) + sec (arcsec (\frac{1}{a} x))) + \frac{sec ^{4} ( arcsec ( \frac{1}{a} x ) ) sin ( arcsec ( \frac{1}{a} x ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arcsec (\frac{1}{a} x)) tan (arcsec (\frac{1}{a} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{a} x)) + sec (arcsec (\frac{1}{a} x))))

Vereinfache

a^{4} (- (\frac{1}{2} sec (arcsec (\frac{1}{a} x)) tan (arcsec (\frac{1}{a} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{a} x)) + sec (arcsec (\frac{1}{a} x))) + \frac{sec ^{4} ( arcsec ( \frac{1}{a} x ) ) sin ( arcsec ( \frac{1}{a} x ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arcsec (\frac{1}{a} x)) tan (arcsec (\frac{1}{a} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{a} x)) + sec (arcsec (\frac{1}{a} x)))) : - \frac{a ^{2} x x ^{2} - a ^{2}}{8} - \frac{a ^{4}}{8} ln (\frac{x ^{2} - a ^{2} + x}{∣ a ∣}) + \frac{1}{4} x^{3} x^{2} - a^{2}

= - \frac{a ^{2} x x ^{2} - a ^{2}}{8} - \frac{a ^{4}}{8} ln (\frac{x ^{2} - a ^{2} + x}{∣ a ∣}) + \frac{1}{4} x^{3} x^{2} - a^{2}

Füge eine Konstante zur Lösung hinzu

= - \frac{a ^{2} x x ^{2} - a ^{2}}{8} - \frac{a ^{4}}{8} ln (\frac{x ^{2} - a ^{2} + x}{∣ a ∣}) + \frac{1}{4} x^{3} x^{2} - a^{2} + C

\int_{- \infty}^{- 1} e^{- 8 t} d t

a re a y = x^{2} - 3 x, x = 0, x = 7, y = 4 x

\frac{\partial}{\partial y} (x y + yz + x z)

x \to (\frac{π}{2}) - lim (tan (x))

\frac{\partial}{\partial x} (x^{m} y^{n})