Question Number 224182 by CrispyXYZ last updated on 24/Aug/25
$$\mathrm{Calculate} \\ $$$${I}=\int\:\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\:\mathrm{sin}\:{x}}\:\mathrm{d}{x} \\ $$
Answered by Frix last updated on 24/Aug/25
$$\int\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}{dx}=\int\frac{\left(\mathrm{1}−\mathrm{sin}\:{x}\right)\mathrm{sin}\:{x}}{\left(\mathrm{1}−\mathrm{sin}\:{x}\right)\left(\mathrm{1}+\mathrm{sin}\:{x}\right)}{dx}= \\ $$$$=\int\frac{\mathrm{sin}\:{x}\:−\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$=\int{dx}−\int\frac{{dx}}{\mathrm{cos}^{\mathrm{2}} \:{x}}+\int\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$={x}−\mathrm{tan}\:{x}\:+\frac{\mathrm{1}}{\mathrm{cos}\:{x}}+{C} \\ $$