y=inu(x)/inx,其中x>0且x≠1,u(x)可导,求dy/dx_作业帮
leica x-u (typ 113)
inu x boku
![=∫[π→0] (π-u)sin(π-u) d(-u) =∫[0→π] (π-u)sinu du =π](http://i.handsetwalls.com/uploads/i_0_1833780786x3766256141_253.jpg)
=∫[π→0] (π-u)sin(π-u) d(-u) =∫[0→π] (π-u)sinu du =π
2条回答 10 分钟前 乐正客|二级 令arcsinx=u,则x=sinu,dx=cosudu 原
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