integral from 0 to pi/2 of cos(x)cos(nx)

解

\int_{0}^{\frac{π}{2}} cos (x) cos (n x) d x

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

解答ステップ

\int_{0}^{\frac{π}{2}} cos (x) cos (n x) d x

三角関数の公式を使用して書き換える

= \int_{0}^{\frac{π}{2}} \frac{cos ( x + n x ) + cos ( 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_{0}^{\frac{π}{2}} cos (x + n x) + cos (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_{0}^{\frac{π}{2}} cos (x + n x) d x + \int_{0}^{\frac{π}{2}} cos (x - n x) d x)

\int_{0}^{\frac{π}{2}} cos (x + n x) d x = \frac{sin ( \frac{π + πn}{2} )}{1 + n}

\int_{0}^{\frac{π}{2}} cos (x - n x) d x = \frac{sin ( \frac{π - πn}{2} )}{1 - n}

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