integral from 0 to 1 of xsin((npix)/2)

解答

\int_{0}^{1} x sin (\frac{nπ x}{2}) d x

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

求解步骤

\int_{0}^{1} x sin (\frac{nπ x}{2}) d x

使用换元积分法

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

提出常数:

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

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

使用分布积分法

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

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

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

化简

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

带入上下限计算后得:

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

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

化简

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