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e-ln-sin-1-x-1-x-2-dx-




Question Number 136929 by liberty last updated on 27/Mar/21
∫ (e^(ln (sin^(−1) x)) /( (√(1−x^2 )))) dx = ?
$$\int\:\frac{\mathrm{e}^{\mathrm{ln}\:\left(\mathrm{sin}^{−\mathrm{1}} \mathrm{x}\right)} }{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\:\mathrm{dx}\:=\:?\: \\ $$
Answered by Dwaipayan Shikari last updated on 27/Mar/21
∫(e^(log(sin^(−1) x)) /( (√(1−x^2 ))))dx     sin^(−1) x=t  ∫e^(log(t)) dt=(t^2 /2)+C=(((sin^(−1) x)^2 )/2)+C
$$\int\frac{{e}^{{log}\left({sin}^{−\mathrm{1}} {x}\right)} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\:\:\:\:\:{sin}^{−\mathrm{1}} {x}={t} \\ $$$$\int{e}^{{log}\left({t}\right)} {dt}=\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+{C}=\frac{\left({sin}^{−\mathrm{1}} {x}\right)^{\mathrm{2}} }{\mathrm{2}}+{C} \\ $$
Answered by liberty last updated on 27/Mar/21
∫ (e^(ln (sin^(−1) (x))) /( (√(1−x^2 )))) dx = ∫ sin^(−1) (x) d(sin^(−1) (x))  = 2(((sin^(−1) (x))/2))^2 + C
$$\int\:\frac{\mathrm{e}^{\mathrm{ln}\:\left(\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)} }{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\:\mathrm{dx}\:=\:\int\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\:\mathrm{d}\left(\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\right) \\ $$$$=\:\mathrm{2}\left(\frac{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{2}}\right)^{\mathrm{2}} +\:\mathrm{C} \\ $$

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