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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∫1dx−∫(x/(√(1−x^2 )))dx  xsin^(−1) x +(1/2)∫((−2x)/(√(1−x^2 )))dx   =xsin^(−1) x+(√(1−x^2 ))+c

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

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

∫f^(−1) (x)dx=xf^(−1) (x)−F○f^(−1) (x)+C  let  f^(−1) (x)=sin^(−1) (x)⇒f(x)=sin(x)  ⇒F(x)=∫sin(x)dx=−cos(x)  ∫sin^(−1) (x)dx=xsin^(−1) (x)+cos(sin^(−1) (x))+C  cos(sin^(−1) (x))=cos(α) [sin^(−1) (x)=α⇒sin(α)=x]  drawing a right triangle with angle α:  sin(α)=x⇒((opp)/(hyp))=(x/1)⇒opp=x and hyp=1  ⇒adj=(√(1−x^2 ))⇒cos(α)=(√(1−x^2 ))  ∫sin^(−1) (x)dx=xsin^(−1) (x)+(√(1−x^2 ))+C

$$\int{f}^{−\mathrm{1}} \left({x}\right){dx}={xf}^{−\mathrm{1}} \left({x}\right)−{F}\circ{f}^{−\mathrm{1}} \left({x}\right)+{C} \\ $$$${let}\:\:{f}^{−\mathrm{1}} \left({x}\right)={sin}^{−\mathrm{1}} \left({x}\right)\Rightarrow{f}\left({x}\right)={sin}\left({x}\right) \\ $$$$\Rightarrow{F}\left({x}\right)=\int{sin}\left({x}\right){dx}=−{cos}\left({x}\right) \\ $$$$\int{sin}^{−\mathrm{1}} \left({x}\right){dx}={xsin}^{−\mathrm{1}} \left({x}\right)+{cos}\left({sin}^{−\mathrm{1}} \left({x}\right)\right)+{C} \\ $$$${cos}\left({sin}^{−\mathrm{1}} \left({x}\right)\right)={cos}\left(\alpha\right)\:\left[{sin}^{−\mathrm{1}} \left({x}\right)=\alpha\Rightarrow{sin}\left(\alpha\right)={x}\right] \\ $$$${drawing}\:{a}\:{right}\:{triangle}\:{with}\:{angle}\:\alpha: \\ $$$${sin}\left(\alpha\right)={x}\Rightarrow\frac{{opp}}{{hyp}}=\frac{{x}}{\mathrm{1}}\Rightarrow{opp}={x}\:{and}\:{hyp}=\mathrm{1} \\ $$$$\Rightarrow{adj}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\Rightarrow{cos}\left(\alpha\right)=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\int{sin}^{−\mathrm{1}} \left({x}\right){dx}={xsin}^{−\mathrm{1}} \left({x}\right)+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{C} \\ $$

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

I=∫sin^(−1) (x)dx  I=∫sin^(−1) (x)+(x/(√(1−x^2 )))dx−∫(x/(√(1−x^2 )))dx  I=∫d(xsin^(−1) (x))−∫(x/(√(1−x^2 )))dx  I=xsin^(−1) (x)+(√(1−x^2 ))+c

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

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