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Question Number 122877 by bemath last updated on 20/Nov/20

$$\int {\left({\sin }^{-1}\left(x\right)\right)}^{2}dx?$$

Commented by liberty last updated on 20/Nov/20

$$letu={\sin }^{-1}\left(x\right)\Rightarrow x=\sin u$$$$\Rightarrow dx=\cos udu$$$$\beta \left(x\right)=\int u^2\cos udu=u^2\sin u-2\int u\sin udu$$$$=u^2\sin u-2(-u\cos u+\int \cos udu)$$$$=u^2\sin u+2u\cos u-2\sin u+c$$$$=(u^2-2)\sin u+2u\cos u+c$$$$=x[{\left({\sin }^{-1}\left(x\right)\right)}^2-2]+{2\sin }^{-1}\left(x\right)\sqrt{1-x^2}+c$$

Answered by mathmax by abdo last updated on 21/Nov/20

$$\mathrm{A}=\int {\left(\mathrm{arcsinx}\right)}^2\mathrm{dx}\Rightarrow \mathrm{A}{=}_{\mathrm{arcsinx}=\mathrm{t}}\int {\mathrm{t}}^2\mathrm{cost}\mathrm{dt}$$$$={\mathrm{t}}^2\mathrm{sint}-\int 2\mathrm{t}\mathrm{sint}\mathrm{dt}={\mathrm{t}}^2\mathrm{sint}-2\int \mathrm{t}\mathrm{sint}\mathrm{dt}$$$$={\mathrm{t}}^2\mathrm{sint}-2(-\mathrm{tcost}+\int \mathrm{cost}\mathrm{dt})$$$$={\mathrm{t}}^2\mathrm{sint}+2\mathrm{t}\mathrm{cost}-2\mathrm{sint}+\mathrm{C}$$$$=\mathrm{x}{\arcsin }^2\mathrm{x}+2\mathrm{arcsinx}\sqrt{1-{\mathrm{t}}^2}-2\mathrm{x}+\mathrm{C}$$

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