integral from 0 to pi of (10+x)cos(nx)

解

\int_{0}^{π} (10 + x) cos (n x) d x

\frac{( - 1 ) ^{n} - 1}{n ^{2}}

解答ステップ

\int_{0}^{π} (10 + x) cos (n x) d x

部分積分を適用する

= [\frac{1}{n} (sin (n x) (10 + x) - n \cdot \int \frac{1}{n} sin (n x) d x)]_{0}^{π}

\int \frac{1}{n} sin (n x) d x = - \frac{1}{n ^{2}} cos (n x)

= [\frac{1}{n} (sin (n x) (10 + x) - n (- \frac{1}{n ^{2}} cos (n x)))]_{0}^{π}

簡素化

[\frac{1}{n} (sin (n x) (10 + x) - n (- \frac{1}{n ^{2}} cos (n x)))]_{0}^{π} : [\frac{1}{n} (sin (n x) (10 + x) + \frac{1}{n} cos (n x))]_{0}^{π}

= [\frac{1}{n} (sin (n x) (10 + x) + \frac{1}{n} cos (n x))]_{0}^{π}

限度を計算する:

(- 1)^{n} \frac{1}{n ^{2}} - \frac{1}{n ^{2}}

= (- 1)^{n} \frac{1}{n ^{2}} - \frac{1}{n ^{2}}

簡素化

= \frac{( - 1 ) ^{n} - 1}{n ^{2}}