integral of xsin^2(7x)

Lösung

\int x sin^{2} (7 x) d x

\frac{1}{392} (98 x^{2} - 14 x sin (14 x) - cos (14 x)) + C

Schritte zur Lösung

\int x sin^{2} (7 x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int \frac{1}{2} x (1 - cos (14 x)) d x

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int x (1 - cos (14 x)) d x

Wende die partielle Integration an

= \frac{1}{2} (x (x - \frac{1}{14} sin (14 x)) - \int x - \frac{1}{14} sin (14 x) d x)

\int x - \frac{1}{14} sin (14 x) d x = \frac{x ^{2}}{2} + \frac{1}{196} cos (14 x)

= \frac{1}{2} (x (x - \frac{1}{14} sin (14 x)) - (\frac{x ^{2}}{2} + \frac{1}{196} cos (14 x)))

Vereinfache

\frac{1}{2} (x (x - \frac{1}{14} sin (14 x)) - (\frac{x ^{2}}{2} + \frac{1}{196} cos (14 x))) : \frac{1}{392} (98 x^{2} - 14 x sin (14 x) - cos (14 x))

= \frac{1}{392} (98 x^{2} - 14 x sin (14 x) - cos (14 x))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{392} (98 x^{2} - 14 x sin (14 x) - cos (14 x)) + C

\int (3 x + 8)^{\frac{2}{3}} d x

ma c l a u r in f (x) = sin (2 x)

\int t^{3} sin (n t) d t

\int 18 x^{2} (2 x^{3} + 3)^{5} d x

\frac{d}{d x} (arccoth (x))