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integral of sqrt(x)(sin(x^{3/2}-1))^2

Lösung

\int x (sin (x^{\frac{3}{2}} - 1))^{2} d x

\frac{1}{6} (2 x^{\frac{3}{2}} - 2 - sin (2 (x^{\frac{3}{2}} - 1))) + C

Schritte zur Lösung

\int x (sin (x^{\frac{3}{2}} - 1))^{2} d x

Vereinfache

= \int x sin^{2} (x^{\frac{3}{2}} - 1) d x

Wende U-Substitution an

= \int \frac{2 sin ^{2} ( u )}{3} d u

Entferne die Konstante:

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

= \frac{2}{3} \cdot \int sin^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \frac{2}{3} \cdot \int \frac{1 - cos ( 2 u )}{2} d u

Entferne die Konstante:

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

= \frac{2}{3} \cdot \frac{1}{2} \cdot \int 1 - cos (2 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

= \frac{2}{3} \cdot \frac{1}{2} (\int 1 d u - \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= \frac{2}{3} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Setze in

u = x^{\frac{3}{2}} - 1

ein

= \frac{2}{3} \cdot \frac{1}{2} (x^{\frac{3}{2}} - 1 - \frac{1}{2} sin (2 (x^{\frac{3}{2}} - 1)))

Vereinfache

\frac{2}{3} \cdot \frac{1}{2} (x^{\frac{3}{2}} - 1 - \frac{1}{2} sin (2 (x^{\frac{3}{2}} - 1))) : \frac{1}{6} (2 x^{\frac{3}{2}} - 2 - sin (2 (x^{\frac{3}{2}} - 1)))

= \frac{1}{6} (2 x^{\frac{3}{2}} - 2 - sin (2 (x^{\frac{3}{2}} - 1)))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} (2 x^{\frac{3}{2}} - 2 - sin (2 (x^{\frac{3}{2}} - 1))) + C

\frac{d}{d x} ((2 x^{2} - 7 x + 6)^{2})

l a pl a ce t r an s f or m 7

f (x) = 6 - x

\int 3 tan (7 x) sec^{2} (7 x) d x

\frac{\partial}{\partial y} (y^{2} - 4 x)