integral from-pi/2 to pi/2 of xsin(nx)

解答

\int_{- \frac{π}{2}}^{\frac{π}{2}} x sin (n x) d x

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

求解步骤

\int_{- \frac{π}{2}}^{\frac{π}{2}} x sin (n x) d x

使用分布积分法

= [\frac{1}{n} (- x cos (n x) - n \cdot \int - \frac{1}{n} cos (n x) d x)]_{- \frac{π}{2}}^{\frac{π}{2}}

\int - \frac{1}{n} cos (n x) d x = - \frac{1}{n ^{2}} sin (n x)

= [\frac{1}{n} (- x cos (n x) - n (- \frac{1}{n ^{2}} sin (n x)))]_{- \frac{π}{2}}^{\frac{π}{2}}

化简

[\frac{1}{n} (- x cos (n x) - n (- \frac{1}{n ^{2}} sin (n x)))]_{- \frac{π}{2}}^{\frac{π}{2}} : [\frac{1}{n} (- x cos (n x) + \frac{1}{n} sin (n x))]_{- \frac{π}{2}}^{\frac{π}{2}}

= [\frac{1}{n} (- x cos (n x) + \frac{1}{n} sin (n x))]_{- \frac{π}{2}}^{\frac{π}{2}}

带入上下限计算后得:

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

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

化简

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

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