integral of sin(ax)cos(bx)

解

\int sin (a x) cos (b x) d x

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

解答ステップ

\int sin (a x) cos (b 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 ( a x + b x ) + sin ( a x - b 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 (a x + b x) + sin (a x - b 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 (a x + b x) d x + \int sin (a x - b x) d x)

\int sin (a x + b x) d x = - \frac{1}{a + b} cos (a x + b x)

\int sin (a x - b x) d x = - \frac{1}{a - b} cos (a x - b x)

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

定数を解答に追加する

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