integral from 0 to 2 of xsin((npix)/4)

解

\int_{0}^{2} x sin (\frac{nπ x}{4}) d x

\frac{16 sin ( \frac{πn}{2} ) - 8 πn cos ( \frac{πn}{2} )}{π ^{2} n ^{2}}

解答ステップ

\int_{0}^{2} x sin (\frac{nπ x}{4}) d x

u置換積分法を適用する

= \int_{0}^{nπ 2} \frac{u sin ( \frac{u}{4} )}{π ^{2} n ^{2}} d u

定数を除く:

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

= \frac{1}{π ^{2} n ^{2}} \cdot \int_{0}^{nπ 2} u sin (\frac{u}{4}) d u

部分積分を適用する

= \frac{1}{π ^{2} n ^{2}} [- 4 u cos (\frac{u}{4}) - \int - 4 cos (\frac{u}{4}) d u]_{0}^{2 πn}

\int - 4 cos (\frac{u}{4}) d u = - 16 sin (\frac{u}{4})

= \frac{1}{π ^{2} n ^{2}} [- 4 u cos (\frac{u}{4}) - (- 16 sin (\frac{u}{4}))]_{0}^{2 πn}

簡素化

= \frac{1}{π ^{2} n ^{2}} [- 4 u cos (\frac{u}{4}) + 16 sin (\frac{u}{4})]_{0}^{2 πn}

限度を計算する:

16 sin (\frac{πn}{2}) - 8 πn cos (\frac{πn}{2})

= \frac{1}{π ^{2} n ^{2}} (16 sin (\frac{πn}{2}) - 8 πn cos (\frac{πn}{2}))

簡素化

= \frac{16 sin ( \frac{πn}{2} ) - 8 πn cos ( \frac{πn}{2} )}{π ^{2} n ^{2}}