integral of sin(6x)cos(7x)

解

\int sin (6 x) cos (7 x) d x

\frac{1}{2} (- \frac{1}{13} cos (13 x) + cos (x)) + C

解答ステップ

\int sin (6 x) cos (7 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 ( 6 x + 7 x ) + sin ( 6 x - 7 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 (6 x + 7 x) + sin (6 x - 7 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 (6 x + 7 x) d x + \int sin (6 x - 7 x) d x)

\int sin (6 x + 7 x) d x = - \frac{1}{13} cos (13 x)

\int sin (6 x - 7 x) d x = cos (x)

= \frac{1}{2} (- \frac{1}{13} cos (13 x) + cos (x))

定数を解答に追加する

= \frac{1}{2} (- \frac{1}{13} cos (13 x) + cos (x)) + C