Question Number 91004 by jagoll last updated on 27/Apr/20
$$\int\:\mathrm{tan}\:\left({arc}\:\mathrm{sin}\:{x}\right)\:{dx} \\ $$
Commented by jagoll last updated on 27/Apr/20
$${let}\:\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\:=\:{y}\:\Rightarrow{x}\:=\:\mathrm{sin}\:{y} \\ $$$${dx}\:=\:\mathrm{cos}\:{y}\:{dy}\: \\ $$$$\mathrm{tan}\:\left(\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\right)\:=\:\mathrm{tan}\:{y}\: \\ $$$$\int\:\mathrm{tan}\:\left({y}\right)\:\mathrm{cos}\:\left({y}\right)\:{dy}\:=\: \\ $$$$\int\:\mathrm{sin}\:\left({y}\right)\:{dy}\:=\:−\mathrm{cos}\:{y}\:+\:{c} \\ $$$$=\:−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+\:{c}\: \\ $$
Commented by MJS last updated on 27/Apr/20
$$\mathrm{tan}\:\mathrm{arcsin}\:{x}\:=\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\mathrm{anyway}\:\mathrm{your}\:\mathrm{path}\:\mathrm{is}\:\mathrm{right} \\ $$
Commented by jagoll last updated on 27/Apr/20
$${waw}\:{you}\:{are}\:{great}\:{sir}... \\ $$$${thank}\:{you}\: \\ $$