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Question Number 101451 by yahyajan last updated on 02/Jul/20

Commented by Dwaipayan Shikari last updated on 02/Jul/20

$${sin}^{-1}x\int 1dx-\int \frac{x}{\sqrt{1-x^2}}dx$$$${xsin}^{-1}x+\frac{1}{2}\int \frac{-2x}{\sqrt{1-x^2}}dx={xsin}^{-1}x+\sqrt{1-x^2}+c$$

Answered by floor(10²Eta[1]) last updated on 02/Jul/20

$$\int f^{-1}\left(x\right)dx={xf}^{-1}\left(x\right)-F\circ f^{-1}\left(x\right)+C$$$$let{f}^{-1}\left(x\right)={sin}^{-1}\left(x\right)\Rightarrow f\left(x\right)=sin\left(x\right)$$$$\Rightarrow F\left(x\right)=\int sin\left(x\right)dx=-cos\left(x\right)$$$$\int {sin}^{-1}\left(x\right)dx={xsin}^{-1}\left(x\right)+cos\left({sin}^{-1}\left(x\right)\right)+C$$$$cos\left({sin}^{-1}\left(x\right)\right)=cos\left(\alpha \right)[{sin}^{-1}\left(x\right)=\alpha \Rightarrow sin\left(\alpha \right)=x]$$$$drawingarighttrianglewithangle\alpha :$$$$sin\left(\alpha \right)=x\Rightarrow \frac{opp}{hyp}=\frac{x}{1}\Rightarrow opp=xandhyp=1$$$$\Rightarrow adj=\sqrt{1-x^2}\Rightarrow cos\left(\alpha \right)=\sqrt{1-x^2}$$$$\int {sin}^{-1}\left(x\right)dx={xsin}^{-1}\left(x\right)+\sqrt{1-x^2}+C$$

Answered by  M±th+et+s last updated on 02/Jul/20

$$I=\int {sin}^{-1}\left(x\right)dx$$$$I=\int {sin}^{-1}\left(x\right)+\frac{x}{\sqrt{1-x^2}}dx-\int \frac{x}{\sqrt{1-x^2}}dx$$$$I=\int d\left({xsin}^{-1}\left(x\right)\right)-\int \frac{x}{\sqrt{1-x^2}}dx$$$$I={xsin}^{-1}\left(x\right)+\sqrt{1-x^2}+c$$

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