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

解

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

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

解答ステップ

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

u置換積分法を適用する

= \int_{0}^{\frac{πn}{2}} \frac{u cos ( u )}{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}{n ^{2}} \cdot \int_{0}^{\frac{πn}{2}} u cos (u) d u

部分積分を適用する

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

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

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

簡素化

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

限度を計算する:

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

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

簡素化

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