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

解

\int \frac{( sin ( 3 x ) - cos ( 2 x ) )}{4} d x

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

解答ステップ

\int \frac{sin ( 3 x ) - cos ( 2 x )}{4} d x

定数を除く:

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

= \frac{1}{4} \cdot \int sin (3 x) - cos (2 x) 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

= \frac{1}{4} (\int sin (3 x) d x - \int cos (2 x) d x)

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

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

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

定数を解答に追加する

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