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受欢迎的微积分 >

integral of xsin(pi/2 x)

解答

\int x sin (\frac{π}{2} x) d x

解答

\frac{4}{π ^{2}} (- \frac{π}{2} x cos (\frac{π x}{2}) + sin (\frac{π x}{2})) + C

求解步骤

\int x sin (\frac{π}{2} x) d x

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \int x sin (\frac{π x}{2}) d x

使用换元积分法

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

提出常数:

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

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

使用分布积分法

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

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

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

u = \frac{π x}{2}

代回

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

化简

\frac{4}{π ^{2}} (- \frac{π x}{2} cos (\frac{π x}{2}) - (- sin (\frac{π x}{2}))) : \frac{4}{π ^{2}} (- \frac{π}{2} x cos (\frac{π x}{2}) + sin (\frac{π x}{2}))

= \frac{4}{π ^{2}} (- \frac{π}{2} x cos (\frac{π x}{2}) + sin (\frac{π x}{2}))

解答补常数

= \frac{4}{π ^{2}} (- \frac{π}{2} x cos (\frac{π x}{2}) + sin (\frac{π x}{2})) + C

作图

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