integral from 1 to 2 of xcos((npix)/2)

解

\int_{1}^{2} x cos (\frac{nπ x}{2}) d x

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

解答ステップ

\int_{1}^{2} x cos (\frac{nπ x}{2}) d x

u置換積分法を適用する

= \int_{πn}^{nπ 2} \frac{u cos ( \frac{u}{2} )}{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_{πn}^{nπ 2} u cos (\frac{u}{2}) d u

u置換積分法を適用する

= \frac{1}{π ^{2} n ^{2}} \cdot \int_{\frac{πn}{2}}^{πn} 4 v cos (v) d v

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 4 \cdot \int_{\frac{πn}{2}}^{πn} v cos (v) d v

部分積分を適用する

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

\int sin (v) d v = - cos (v)

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

簡素化

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

= \frac{4}{π ^{2} n ^{2}} [v sin (v) + cos (v)]_{\frac{πn}{2}}^{πn}

限度を計算する:

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

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

簡素化

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