integral from 0 to 1 of-sin((npix)/2)

解答

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

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

求解步骤

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

提出常数:

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

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

\frac{nπ x}{2} = \frac{nπ}{2} x

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

使用换元积分法

= - \int_{0}^{\frac{πn}{2}} \frac{2 sin ( u )}{πn} d u

提出常数:

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

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

使用积分常见定则:

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

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

带入上下限计算后得:

- cos (\frac{πn}{2}) + 1

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

化简

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

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