integral of (2-3x)cos(4x)

Lösung

\int (2 - 3 x) cos (4 x) d x

\frac{1}{2} sin (4 x) - \frac{3}{16} (4 x sin (4 x) + cos (4 x)) + C

Schritte zur Lösung

\int (2 - 3 x) cos (4 x) d x

Multipliziere aus

(2 - 3 x) cos (4 x) : 2 cos (4 x) - 3 x cos (4 x)

= \int 2 cos (4 x) - 3 x cos (4 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 2 cos (4 x) d x - \int 3 x cos (4 x) d x

\int 2 cos (4 x) d x = \frac{1}{2} sin (4 x)

\int 3 x cos (4 x) d x = \frac{3}{16} (4 x sin (4 x) + cos (4 x))

= \frac{1}{2} sin (4 x) - \frac{3}{16} (4 x sin (4 x) + cos (4 x))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} sin (4 x) - \frac{3}{16} (4 x sin (4 x) + cos (4 x)) + C

\frac{\partial}{\partial x} (x^{2} + y^{2})

\int sech (x) d x

\frac{d}{d x} (4 3 x^{7} - \frac{1}{x ^{2}} + 2 x)

t an g e n t 7 x - 7 x

\frac{d}{d x} ((\frac{4 x + 1}{2 - 5 x})^{\frac{1}{3}})