integral from 0 to pi of xsin((nx)/2)

解

\int_{0}^{π} x sin (\frac{n x}{2}) d x

\frac{4 sin ( \frac{πn}{2} ) - 2 πn cos ( \frac{πn}{2} )}{n ^{2}}

解答ステップ

\int_{0}^{π} x sin (\frac{n x}{2}) d x

部分積分を適用する

= [\frac{1}{n} (- 2 x cos (\frac{n x}{2}) - n \cdot \int - \frac{2}{n} cos (\frac{n x}{2}) d x)]_{0}^{π}

\int - \frac{2}{n} cos (\frac{n x}{2}) d x = - \frac{4}{n ^{2}} sin (\frac{n x}{2})

= [\frac{1}{n} (- 2 x cos (\frac{n x}{2}) - n (- \frac{4}{n ^{2}} sin (\frac{n x}{2})))]_{0}^{π}

簡素化

[\frac{1}{n} (- 2 x cos (\frac{n x}{2}) - n (- \frac{4}{n ^{2}} sin (\frac{n x}{2})))]_{0}^{π} : [\frac{1}{n} (- 2 x cos (\frac{n x}{2}) + \frac{4}{n} sin (\frac{n x}{2}))]_{0}^{π}

= [\frac{1}{n} (- 2 x cos (\frac{n x}{2}) + \frac{4}{n} sin (\frac{n x}{2}))]_{0}^{π}

限度を計算する:

\frac{1}{n} (\frac{4}{n} sin (\frac{πn}{2}) - 2 π cos (\frac{πn}{2}))

= \frac{1}{n} (\frac{4}{n} sin (\frac{πn}{2}) - 2 π cos (\frac{πn}{2}))

簡素化

= \frac{4 sin ( \frac{πn}{2} ) - 2 πn cos ( \frac{πn}{2} )}{n ^{2}}