integral of+cos(9x)sin(5x)

解

\int + cos (9 x) sin (5 x) d x

\frac{1}{2} (- \frac{1}{14} cos (14 x) + \frac{1}{4} cos (4 x)) + C

解答ステップ

\int cos (9 x) sin (5 x) d x

次の恒等を使用する:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int \frac{sin ( 5 x + 9 x ) + sin ( 5 x - 9 x )}{2} d x

定数を除く:

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

= \frac{1}{2} \cdot \int sin (5 x + 9 x) + sin (5 x - 9 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}{2} (\int sin (5 x + 9 x) d x + \int sin (5 x - 9 x) d x)

\int sin (5 x + 9 x) d x = - \frac{1}{14} cos (14 x)

\int sin (5 x - 9 x) d x = \frac{1}{4} cos (4 x)

= \frac{1}{2} (- \frac{1}{14} cos (14 x) + \frac{1}{4} cos (4 x))

定数を解答に追加する

= \frac{1}{2} (- \frac{1}{14} cos (14 x) + \frac{1}{4} cos (4 x)) + C