integral of sin(9x)cos(2x)

解

\int sin (9 x) cos (2 x) d x

\frac{1}{2} (- \frac{1}{11} cos (11 x) - \frac{1}{7} cos (7 x)) + C

解答ステップ

\int sin (9 x) cos (2 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 ( 9 x + 2 x ) + sin ( 9 x - 2 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 (9 x + 2 x) + sin (9 x - 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}{2} (\int sin (9 x + 2 x) d x + \int sin (9 x - 2 x) d x)

\int sin (9 x + 2 x) d x = - \frac{1}{11} cos (11 x)

\int sin (9 x - 2 x) d x = - \frac{1}{7} cos (7 x)

= \frac{1}{2} (- \frac{1}{11} cos (11 x) - \frac{1}{7} cos (7 x))

定数を解答に追加する

= \frac{1}{2} (- \frac{1}{11} cos (11 x) - \frac{1}{7} cos (7 x)) + C