integral of cos(10x)sin(4x+1)

解

\int cos (10 x) sin (4 x + 1) d x

\frac{1}{2} (- \frac{1}{14} cos (14 x + 1) + \frac{1}{6} cos (- 6 x + 1)) + C

解答ステップ

\int cos (10 x) sin (4 x + 1) 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 ( 4 x + 1 + 10 x ) + sin ( 4 x + 1 - 10 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 (4 x + 1 + 10 x) + sin (4 x + 1 - 10 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 (4 x + 1 + 10 x) d x + \int sin (4 x + 1 - 10 x) d x)

\int sin (4 x + 1 + 10 x) d x = - \frac{1}{14} cos (14 x + 1)

\int sin (4 x + 1 - 10 x) d x = \frac{1}{6} cos (- 6 x + 1)

= \frac{1}{2} (- \frac{1}{14} cos (14 x + 1) + \frac{1}{6} cos (- 6 x + 1))

定数を解答に追加する

= \frac{1}{2} (- \frac{1}{14} cos (14 x + 1) + \frac{1}{6} cos (- 6 x + 1)) + C