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Question Number 172608 by Mikenice last updated on 29/Jun/22
Answered by FelipeLz last updated on 03/Jul/22
f(x)=x−sin(x)∫xcos(x)−sin(x)[x−sin(x)]2dx∫xcos(x)−x+x−sin(x)[x−sin(x)]2dx∫[x−sin(x)]−x[1−cos(x)][x−sin(x)]2dx∫f(x)−xf′(x)[f(x)]2dx=xf(x)+c=xx−sin(x)+c
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