integral from 0 to 1 of-sin((npix)/2)

解

\int_{0}^{1} - sin (\frac{nπ x}{2}) d x

- \frac{2 ( - cos ( \frac{πn}{2} ) + 1 )}{πn}

解答ステップ

\int_{0}^{1} - sin (\frac{nπ x}{2}) d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int_{0}^{1} sin (\frac{nπ x}{2}) d x

\frac{nπ x}{2} = \frac{nπ}{2} x

= - \int_{0}^{1} sin (\frac{nπ}{2} x) d x

u置換積分法を適用する

= - \int_{0}^{\frac{πn}{2}} \frac{2 sin ( u )}{πn} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{2}{πn} \cdot \int_{0}^{\frac{πn}{2}} sin (u) d u

共通積分を使用する:

\int sin (u) d u = - cos (u)

= - \frac{2}{πn} [- cos (u)]_{0}^{\frac{πn}{2}}

限度を計算する:

- cos (\frac{πn}{2}) + 1

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

簡素化

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