integral from 0 to 1 of cos(ax)cos(bx)

解

\int_{0}^{1} cos (a x) cos (b x) d x

\frac{1}{2} (\frac{sin ( a + b )}{a + b} + \frac{sin ( a - b )}{a - b})

解答ステップ

\int_{0}^{1} cos (a x) cos (b x) d x

三角関数の公式を使用して書き換える

= \int_{0}^{1} \frac{cos ( a x + b x ) + cos ( 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_{0}^{1} cos (a x + b x) + cos (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_{0}^{1} cos (a x + b x) d x + \int_{0}^{1} cos (a x - b x) d x)

\int_{0}^{1} cos (a x + b x) d x = \frac{sin ( a + b )}{a + b}

\int_{0}^{1} cos (a x - b x) d x = \frac{sin ( a - b )}{a - b}

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