integral of 1/2 sin(7y)-1/2 sin(3y)

Lösung

\int \frac{1}{2} sin (7 y) - \frac{1}{2} sin (3 y) d y

- \frac{1}{14} cos (7 y) + \frac{1}{6} cos (3 y) + C

Schritte zur Lösung

\int \frac{1}{2} sin (7 y) - \frac{1}{2} sin (3 y) d y

Wende die Summenregel an:

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

= \int \frac{1}{2} sin (7 y) d y - \int \frac{1}{2} sin (3 y) d y

\int \frac{1}{2} sin (7 y) d y = - \frac{1}{14} cos (7 y)

\int \frac{1}{2} sin (3 y) d y = - \frac{1}{6} cos (3 y)

= - \frac{1}{14} cos (7 y) - (- \frac{1}{6} cos (3 y))

Vereinfache

= - \frac{1}{14} cos (7 y) + \frac{1}{6} cos (3 y)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{14} cos (7 y) + \frac{1}{6} cos (3 y) + C

\int x (- e^{- x}) d x

\int t - 3 d t

\int \frac{1}{x ^{2} - 7} d x

\int 5^{2 x - 1} - 2^{3 x} d x

\int (5 - 2 x)^{7} d x