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受欢迎的微积分 >

integral of-(2ycos((piy)/2))/(pi)

解答

\int - \frac{2 y cos ( \frac{π y}{2} )}{π} d y

解答

- \frac{8}{π ^{3}} (\frac{π}{2} y sin (\frac{π y}{2}) + cos (\frac{π y}{2})) + C

求解步骤

\int - \frac{2 y cos ( \frac{π y}{2} )}{π} d y

提出常数:

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

= - \frac{2}{π} \cdot \int y cos (\frac{π y}{2}) d y

使用换元积分法

= - \frac{2}{π} \cdot \int \frac{4 u cos ( u )}{π ^{2}} d u

提出常数:

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

= - \frac{2}{π} \cdot \frac{4}{π ^{2}} \cdot \int u cos (u) d u

使用分布积分法

= - \frac{2}{π} \cdot \frac{4}{π ^{2}} (u sin (u) - \int sin (u) d u)

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

= - \frac{2}{π} \cdot \frac{4}{π ^{2}} (u sin (u) - (- cos (u)))

u = \frac{π y}{2}

代回

= - \frac{2}{π} \cdot \frac{4}{π ^{2}} (\frac{π y}{2} sin (\frac{π y}{2}) - (- cos (\frac{π y}{2})))

化简

- \frac{2}{π} \cdot \frac{4}{π ^{2}} (\frac{π y}{2} sin (\frac{π y}{2}) - (- cos (\frac{π y}{2}))) : - \frac{8}{π ^{3}} (\frac{π}{2} y sin (\frac{π y}{2}) + cos (\frac{π y}{2}))

= - \frac{8}{π ^{3}} (\frac{π}{2} y sin (\frac{π y}{2}) + cos (\frac{π y}{2}))

解答补常数

= - \frac{8}{π ^{3}} (\frac{π}{2} y sin (\frac{π y}{2}) + cos (\frac{π y}{2})) + C

作图

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