integral of (3x^2-2x-5)sin(2x)

Lösung

\int (3 x^{2} - 2 x - 5) sin (2 x) d x

- \frac{3}{2} x^{2} cos (2 x) + \frac{3}{2} x sin (2 x) + x cos (2 x) + \frac{13}{4} cos (2 x) - \frac{1}{2} sin (2 x) + C

Schritte zur Lösung

\int (3 x^{2} - 2 x - 5) sin (2 x) d x

Multipliziere aus

(3 x^{2} - 2 x - 5) sin (2 x) : 3 x^{2} sin (2 x) - 2 x sin (2 x) - 5 sin (2 x)

= \int 3 x^{2} sin (2 x) - 2 x sin (2 x) - 5 sin (2 x) d x

Wende die Summenregel an:

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

= \int 3 x^{2} sin (2 x) d x - \int 2 x sin (2 x) d x - \int 5 sin (2 x) d x

\int 3 x^{2} sin (2 x) d x = 3 (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x)))

\int 2 x sin (2 x) d x = - x cos (2 x) + \frac{1}{2} sin (2 x)

\int 5 sin (2 x) d x = - \frac{5}{2} cos (2 x)

= 3 (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))) - (- x cos (2 x) + \frac{1}{2} sin (2 x)) - (- \frac{5}{2} cos (2 x))

Vereinfache

3 (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))) - (- x cos (2 x) + \frac{1}{2} sin (2 x)) - (- \frac{5}{2} cos (2 x)) : - \frac{3}{2} x^{2} cos (2 x) + \frac{3}{2} x sin (2 x) + x cos (2 x) + \frac{13}{4} cos (2 x) - \frac{1}{2} sin (2 x)

= - \frac{3}{2} x^{2} cos (2 x) + \frac{3}{2} x sin (2 x) + x cos (2 x) + \frac{13}{4} cos (2 x) - \frac{1}{2} sin (2 x)

Füge eine Konstante zur Lösung hinzu

= - \frac{3}{2} x^{2} cos (2 x) + \frac{3}{2} x sin (2 x) + x cos (2 x) + \frac{13}{4} cos (2 x) - \frac{1}{2} sin (2 x) + C

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

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