2. 计算定积分 ∫π/2到0 xcosxdx

Question

如题所述

Answer 1

∫π/2到0 xcosxdx
原式=∫π/2到0 xdsinx
=xsinx-∫π/2到0 sinxdx
=xsinx+∫π/2到0 dcosx
=(xsinx+cosx)|π/2到0
=[π/2*sin(π/2)+cos(π/2)]-[0*sin0+cos0]
=π/2-1

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Answer 2

xsinx的导数是多少？
(xsinx)'=xcos+sinx 那么就把题目中的积分构造一个xsinx吧！
∫xcosxdx
=∫（xcosx+sinx）dx-∫sinxdx
=xsinx+cosx
所以答案就是
（π/2*1+0）-（0+1）=π/2-1本回答被提问者采纳

Answer 3

原式= ∫π/2到0 xcosxdx
= ∫π/2到0 xdsinx
=x*sinx|（π/2,0）- ∫π/2到0 sinxdx
=π/2+cosx|(π/2,0)
=π/2-1

Answer 4

∫(0->π/2) xcosxdx
=∫(0->π/2) xdsinx
= [xsinx](0->π/2) -∫(0->π/2) sinxdx
=π/2 +[cosx](0->π/2)
=π/2 -1