integral of 6sin(3x)-cos(2x)

Lösung

\int 6 sin (3 x) - cos (2 x) d x

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

Schritte zur Lösung

\int 6 sin (3 x) - cos (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 6 sin (3 x) d x - \int cos (2 x) d x

\int 6 sin (3 x) d x = - 2 cos (3 x)

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d x} (ln (sin (x^{2}) + 1))

\int 4 x^{6} - 2 x^{3} + 7 x - 4 d x

\int_{2}^{5} \frac{1}{x ^{2}} d x

\int_{0}^{2} \frac{( 4 x + 3 )}{x ^{4}} d x

\frac{d}{d x} (a sin^{3} (\frac{x}{3}))