Question Number 159870 by tounghoungko last updated on 21/Nov/21
$$\:\:\int\:\frac{\mathrm{1}−\mathrm{cot}\:^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{sin}\:{x}}\:{dx}\:=? \\ $$
Answered by Ar Brandon last updated on 21/Nov/21
$${I}=\int\frac{\mathrm{1}−\mathrm{cot}^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{sin}{x}}{dx}=\int\frac{\left(\mathrm{1}−\mathrm{cot}^{\mathrm{2}} {x}\right)\left(\mathrm{1}−\mathrm{sin}{x}\right)}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$\:\:\:=\int\frac{\mathrm{1}−\mathrm{sin}{x}−\mathrm{cot}^{\mathrm{2}} {x}+\mathrm{cot}^{\mathrm{2}} {x}\mathrm{sin}{x}}{\mathrm{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\:\:\:=\int\left(\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} {x}}−\frac{\mathrm{sin}{x}}{\mathrm{cos}^{\mathrm{2}} {x}}−\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} {x}}+\frac{\mathrm{1}}{\mathrm{sin}{x}}\right){dx} \\ $$$$\:\:\:=\mathrm{tan}{x}−\frac{\mathrm{1}}{\mathrm{cos}{x}}+\mathrm{cot}{x}+\mathrm{ln}\mid\mathrm{cosec}{x}−\mathrm{cot}{x}\mid+{C} \\ $$$$\:\:\:=\frac{\mathrm{sin}{x}−\mathrm{1}}{\mathrm{cos}{x}}+\frac{\mathrm{cos}{x}}{\mathrm{sin}{x}}+\mathrm{ln}\mid\frac{\mathrm{1}−\mathrm{cos}{x}}{\mathrm{sin}{x}}\mid+{C} \\ $$
Commented by tounghoungko last updated on 22/Nov/21
$${yes}…{very}\:{simple} \\ $$