integral from 0 to pi of 1/n cos(nx)

解

\int_{0}^{π} \frac{1}{n} cos (n x) d x

0

解答ステップ

\int_{0}^{π} \frac{1}{n} cos (n x) d x

定数を除く:

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

= \frac{1}{n} \cdot \int_{0}^{π} cos (n x) d x

u置換積分法を適用する

= \frac{1}{n} \cdot \int_{0}^{nπ} \frac{cos ( u )}{n} d u

定数を除く:

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

= \frac{1}{n} \cdot \frac{1}{n} \cdot \int_{0}^{nπ} cos (u) d u

共通積分を使用する:

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

= \frac{1}{n} \cdot \frac{1}{n} [sin (u)]_{0}^{nπ}

簡素化

\frac{1}{n} \cdot \frac{1}{n} [sin (u)]_{0}^{nπ} : \frac{1}{n ^{2}} [sin (u)]_{0}^{πn}

= \frac{1}{n ^{2}} [sin (u)]_{0}^{πn}

限度を計算する:

0

= \frac{1}{n ^{2}} \cdot 0

簡素化

= 0