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Question Number 127110 by benjo_mathlover last updated on 26/Dec/20

  ∫ (arcsin x)^2  dx =?

$$\:\:\int\:\left(\mathrm{arcsin}\:{x}\right)^{\mathrm{2}} \:{dx}\:=? \\ $$

Answered by liberty last updated on 27/Dec/20

 letting arcsin x = ℓ ⇒x = sin ℓ ∧ dx = cos ℓ dℓ  I= ∫ ℓ^2 cos ℓ dℓ = ℓ^2 sin ℓ +2ℓ cos ℓ−2sin ℓ + c   = x (arcsin x)^2 +2(√(1−x^2 )) arcsin x −2x + c

$$\:{letting}\:\mathrm{arcsin}\:{x}\:=\:\ell\:\Rightarrow{x}\:=\:\mathrm{sin}\:\ell\:\wedge\:{dx}\:=\:\mathrm{cos}\:\ell\:{d}\ell \\ $$$${I}=\:\int\:\ell^{\mathrm{2}} \mathrm{cos}\:\ell\:{d}\ell\:=\:\ell^{\mathrm{2}} \mathrm{sin}\:\ell\:+\mathrm{2}\ell\:\mathrm{cos}\:\ell−\mathrm{2sin}\:\ell\:+\:{c} \\ $$$$\:=\:{x}\:\left(\mathrm{arcsin}\:{x}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\mathrm{arcsin}\:{x}\:−\mathrm{2}{x}\:+\:{c} \\ $$$$ \\ $$

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