integral of cos(x)sin(kx)

解

\int cos (x) sin (k x) d x

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

解答ステップ

\int cos (x) sin (k 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 ( k x + x ) + sin ( k x - 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 (k x + x) + sin (k x - 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 (k x + x) d x + \int sin (k x - x) d x)

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

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

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

定数を解答に追加する

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