integral from-pi to pi of cos(mx)cos(nx)

解

\int_{- π}^{π} cos (m x) cos (n x) d x

\frac{1}{m + n} sin (πm + πn) + \frac{sin ( πm - πn ) - sin ( - πm + πn )}{2 ( m - n )}

解答ステップ

\int_{- π}^{π} cos (m x) cos (n x) d x

三角関数の公式を使用して書き換える

= \int_{- π}^{π} \frac{cos ( m x + n x ) + cos ( m x - n 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_{- π}^{π} cos (m x + n x) + cos (m x - n 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_{- π}^{π} cos (m x + n x) d x + \int_{- π}^{π} cos (m x - n x) d x)

\int_{- π}^{π} cos (m x + n x) d x = \frac{2 sin ( πm + πn )}{m + n}

\int_{- π}^{π} cos (m x - n x) d x = \frac{sin ( πm - πn ) - sin ( - πm + πn )}{m - n}

= \frac{1}{2} (\frac{2 sin ( πm + πn )}{m + n} + \frac{sin ( πm - πn ) - sin ( - πm + πn )}{m - n})

簡素化

\frac{1}{2} (\frac{2 sin ( πm + πn )}{m + n} + \frac{sin ( πm - πn ) - sin ( - πm + πn )}{m - n}) : \frac{1}{m + n} sin (πm + πn) + \frac{sin ( πm - πn ) - sin ( - πm + πn )}{2 ( m - n )}

= \frac{1}{m + n} sin (πm + πn) + \frac{sin ( πm - πn ) - sin ( - πm + πn )}{2 ( m - n )}