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nafanya2014

Замена переменной:

u=arcsin2x\ \ du=(arcsin2x)`dx\ \ du=frac{(2x)`dx}{sqrt{1-(2x)^2} } \ \ du=frac{2dx}{sqrt{1-4x^2} }

intfrac{dx}{sqrt{1-4x^2}arsin^3(2x) }dx =frac{1}{2} int arsin^{-3}(2x)cdot frac{2dx}{sqrt{1-4x^2}} =frac{1}{2} int arsin^{-3}(2x) d(arcsin(2x))=frac{1}{2}cdot frac{arsin^{-2}(2x)}{(-2)}+C=-frac{1}{4arcsin^2(2x)}  +C

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Miroslava227

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