Question Number 177775 by depressiveshrek last updated on 08/Oct/22
$$\int\frac{\mathrm{sin2}{x}}{\:\sqrt{\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{1}}}{dx} \\ $$
Answered by Ar Brandon last updated on 08/Oct/22
$${I}=\int\frac{\mathrm{sin2}{x}}{\:\sqrt{\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{1}}}{dx}=\int\frac{\mathrm{2sin}{x}\mathrm{cos}{x}}{\:\sqrt{\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{1}}}{dx}\:,\:{c}=\mathrm{cos}{x} \\ $$$$\:\:\:=−\int\frac{\mathrm{2}{c}}{\:\sqrt{{c}^{\mathrm{4}} +\mathrm{1}}}{dc}=−\int\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{dt}\:,\:{t}={c}^{\mathrm{2}} \\ $$$$\:\:\:=−\mathrm{argsh}\left({t}\right)+{C}=−\mathrm{argsh}\left({c}^{\mathrm{2}} \right)+{C} \\ $$$$\:\:\:=−\mathrm{argsh}\left(\mathrm{cos}^{\mathrm{2}} {x}\right)+{C} \\ $$$$\:\:\:=−\mathrm{ln}\left(\mathrm{cos}^{\mathrm{2}} {x}+\sqrt{\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{1}}\right)+{C} \\ $$$$\:\:\:=\mathrm{ln}\mid\frac{{k}}{\mathrm{cos}^{\mathrm{2}} {x}+\sqrt{\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{1}}}\mid \\ $$