integral of sqrt(36x^2-144)

Lösung

\int 36 x^{2} - 144 d x

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

Schritte zur Lösung

\int 36 x^{2} - 144 d x

Trigonometrische Substitution anwenden

= \int 24 tan^{2} (u) sec (u) d u

Entferne die Konstante:

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

= 24 \cdot \int tan^{2} (u) sec (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 24 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Multipliziere aus

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

= 24 \cdot \int - sec (u) + sec^{3} (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

= 24 (- \int sec (u) d u + \int sec^{3} (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

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

= 24 (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = arcsec (\frac{6}{12} x)

ein

= 24 (- ln tan (arcsec (\frac{6}{12} x)) + sec (arcsec (\frac{6}{12} x)) + \frac{1}{2} sec (arcsec (\frac{6}{12} x)) tan (arcsec (\frac{6}{12} x)) + \frac{1}{2} ln tan (arcsec (\frac{6}{12} x)) + sec (arcsec (\frac{6}{12} x)))

Vereinfache

24 (- ln tan (arcsec (\frac{6}{12} x)) + sec (arcsec (\frac{6}{12} x)) + \frac{1}{2} sec (arcsec (\frac{6}{12} x)) tan (arcsec (\frac{6}{12} x)) + \frac{1}{2} ln tan (arcsec (\frac{6}{12} x)) + sec (arcsec (\frac{6}{12} x))) : 24 (\frac{1}{8} x x^{2} - 4 - \frac{1}{2} ln (\frac{x + x ^{2} - 4}{2}))

= 24 (\frac{1}{8} x x^{2} - 4 - \frac{1}{2} ln (\frac{x + x ^{2} - 4}{2}))

Füge eine Konstante zur Lösung hinzu

= 24 (\frac{1}{8} x x^{2} - 4 - \frac{1}{2} ln (\frac{x + x ^{2} - 4}{2})) + C

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