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

解答

\int_{\frac{1}{2}}^{1} sin (nπ x) d x

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

求解步骤

\int_{\frac{1}{2}}^{1} sin (nπ x) d x

使用换元积分法

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

提出常数:

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

= \frac{1}{πn} \cdot \int_{\frac{πn}{2}}^{πn} sin (u) d u

使用积分常见定则:

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

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

带入上下限计算后得:

- (- 1)^{n} + cos (\frac{πn}{2})

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

化简

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

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