integral of (2cos(x)+4sin(x)+6)

解

\int (2 cos (x) + 4 sin (x) + 6) d x

2 sin (x) - 4 cos (x) + 6 x + C

解答ステップ

\int 2 cos (x) + 4 sin (x) + 6 d x

総和規則を適用する:

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

= \int 2 cos (x) d x + \int 4 sin (x) d x + \int 6 d x

\int 2 cos (x) d x = 2 sin (x)

\int 4 sin (x) d x = - 4 cos (x)

\int 6 d x = 6 x

= 2 sin (x) - 4 cos (x) + 6 x

定数を解答に追加する

= 2 sin (x) - 4 cos (x) + 6 x + C