integral from-pi to pi of (1+x)cos(nx)

解

\int_{- π}^{π} (1 + x) cos (n x) d x

0

解答ステップ

\int_{- π}^{π} (1 + x) cos (n x) d x

部分積分を適用する

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

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

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

簡素化

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

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

限度を計算する:

0

= 0