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

解

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

- \frac{4 sin ( \frac{πn}{2} )}{πn}

解答ステップ

\int_{- 1}^{0} - 2 cos (\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

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

u置換積分法を適用する

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

定数を除く:

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

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

共通積分を使用する:

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

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

簡素化

- 2 \cdot \frac{2}{πn} [sin (u)]_{- \frac{πn}{2}}^{0} : - \frac{4}{πn} [sin (u)]_{- \frac{πn}{2}}^{0}

= - \frac{4}{πn} [sin (u)]_{- \frac{πn}{2}}^{0}

限度を計算する:

sin (\frac{πn}{2})

= - \frac{4}{πn} sin (\frac{πn}{2})

簡素化

= - \frac{4 sin ( \frac{πn}{2} )}{πn}