integral from 0 to 4 of xcos((npix)/4)

解

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

\frac{16 ( - 1 ) ^{n} - 16}{π ^{2} n ^{2}}

解答ステップ

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

u置換積分法を適用する

= \int_{0}^{nπ 4} \frac{u cos ( \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 4} u cos (\frac{u}{4}) d u

部分積分を適用する

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

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

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

簡素化

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

限度を計算する:

16 (- 1)^{n} - 16

= \frac{1}{π ^{2} n ^{2}} (16 (- 1)^{n} - 16)

簡素化

= \frac{16 ( - 1 ) ^{n} - 16}{π ^{2} n ^{2}}