arcsinx-2-dx-

Question Number 91996 by student work last updated on 04/May/20

$$\int\left(\mathrm{arcsinx}\right)^{\mathrm{2}} \mathrm{dx}=? \\ $$

Commented by jagoll last updated on 04/May/20

let sin^(−1) (x) = t x = sin t ⇒dx = cos t dt ∫ t^2 cos t dt = t^2 sin t +2t cos t −2sin t + c = x(sin^(−1) (x))^2 +2(√(1−x^2 )) (sin^(−1) (x))−2x +c

$$\mathrm{let}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\:=\:\mathrm{t}\: \\ $$$$\mathrm{x}\:=\:\mathrm{sin}\:\mathrm{t}\:\Rightarrow\mathrm{dx}\:=\:\mathrm{cos}\:\mathrm{t}\:\mathrm{dt}\: \\ $$$$\int\:\mathrm{t}^{\mathrm{2}} \:\mathrm{cos}\:\mathrm{t}\:\mathrm{dt}\:=\: \\ $$$$\mathrm{t}^{\mathrm{2}} \mathrm{sin}\:\mathrm{t}\:+\mathrm{2t}\:\mathrm{cos}\:\mathrm{t}\:−\mathrm{2sin}\:\mathrm{t}\:+\:\mathrm{c} \\ $$$$=\:\mathrm{x}\left(\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} \:+\mathrm{2}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\left(\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)−\mathrm{2x}\:+\mathrm{c} \\ $$

Commented by jagoll last updated on 04/May/20

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arcsinx-2-dx-

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