Задача 36547 ...

vk173500291

2019-04-29 10:51:48

∫arcsin ⁡2xdx

sova

2019-04-29 11:01:27

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По частям:∫udv=u·v– ∫ vdu
u=arcsin2x
du=(2x)`dx/sqrt(1-(2x)^2)
du=2/sqrt(1-4x^2)

dv=dx
v= ∫ dx=x

∫arcsin ⁡2xdx=x*arcsin2x- ∫ x*( 2dx)/sqrt(1-4x^2)=

=x*arcsin2x- ∫ 2xdx/sqrt(1-4x^2)=

=x*arcsin2x+ (1/4) ∫ ( -8x)dx/sqrt(1-4x^2)=

=x*arcsin2x+(1/4) ∫(1-4x^2)^(-1/2)d(1-4x^2)=

=x*arcsin2x+(1/4) *(1-4x^2)^(1/2)/(1/2) + C


= [b]x*arcsin2x +(1/2)sqrt(1-4x^2) +C[/b]