integral of sin(3x)cos(6x)

解

\int sin (3 x) cos (6 x) d x

\frac{1}{2} (- \frac{1}{9} cos (9 x) + \frac{1}{3} cos (3 x)) + C

解答ステップ

\int sin (3 x) cos (6 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 ( 3 x + 6 x ) + sin ( 3 x - 6 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 (3 x + 6 x) + sin (3 x - 6 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 (3 x + 6 x) d x + \int sin (3 x - 6 x) d x)

\int sin (3 x + 6 x) d x = - \frac{1}{9} cos (9 x)

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

= \frac{1}{2} (- \frac{1}{9} cos (9 x) + \frac{1}{3} cos (3 x))

定数を解答に追加する

= \frac{1}{2} (- \frac{1}{9} cos (9 x) + \frac{1}{3} cos (3 x)) + C