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受欢迎的微积分 >

integral from 0 to 1 of sin^4(1/2 pix)

解答

\int_{0}^{1} sin^{4} (\frac{1}{2} π x) d x

解答

\frac{3}{8}

+1

十进制

0.375

求解步骤

\int_{0}^{1} sin^{4} (\frac{1}{2} π x) d x

乘

\frac{1}{2} π x : \frac{π}{2} x

= \int_{0}^{1} sin^{4} (\frac{π}{2} x) d x

使用换元积分法

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

提出常数:

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

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

使用降次积分法

= \frac{2}{π} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) d u)

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

= \frac{2}{π} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{π}{4})

化简

\frac{2}{π} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{π}{4}) : - \frac{2}{π} [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{\frac{π}{2}} + \frac{3}{8}

= - \frac{2}{π} [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{\frac{π}{2}} + \frac{3}{8}

带入上下限计算后得:

0

= - \frac{2}{π} \cdot 0 + \frac{3}{8}

化简

= \frac{3}{8}

作图

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