integral from 0 to 1/2 of xcos(pikx)

解答

\int_{0}^{\frac{1}{2}} x cos (πk x) d x

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

求解步骤

\int_{0}^{\frac{1}{2}} x cos (πk x) d x

使用换元积分法

= \int_{0}^{\frac{πk}{2}} \frac{u cos ( u )}{π ^{2} k ^{2}} d u

提出常数:

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

= \frac{1}{π ^{2} k ^{2}} \cdot \int_{0}^{\frac{πk}{2}} u cos (u) d u

使用分布积分法

= \frac{1}{π ^{2} k ^{2}} [u sin (u) - \int sin (u) d u]_{0}^{\frac{πk}{2}}

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

= \frac{1}{π ^{2} k ^{2}} [u sin (u) - (- cos (u))]_{0}^{\frac{πk}{2}}

化简

= \frac{1}{π ^{2} k ^{2}} [u sin (u) + cos (u)]_{0}^{\frac{πk}{2}}

带入上下限计算后得:

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

= \frac{1}{π ^{2} k ^{2}} (\frac{πk}{2} sin (\frac{πk}{2}) + cos (\frac{πk}{2}) - 1)

化简

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

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