Question Number 173296 by DAVONG last updated on 09/Jul/22
$$\mathrm{I}=\int\frac{\left(\mathrm{cosx}−\mathrm{sinx}\right)\left(\mathrm{1}+\mathrm{sin2x}\right)}{\left(\mathrm{sinx}+\mathrm{cosx}\right)}\mathrm{dx}=\: \\ $$
Answered by Jamshidbek last updated on 09/Jul/22
$$\int\frac{\left(\mathrm{cosx}−\mathrm{sinx}\right)\left(\mathrm{sinx}+\mathrm{cosx}\right)^{\mathrm{2}} }{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}=\int\left(\mathrm{cosx}−\mathrm{sinx}\right)\left(\mathrm{cosx}+\mathrm{sinx}\right)\mathrm{dx}= \\ $$$$=\int\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{sin}^{\mathrm{2}} \mathrm{xdx}=\int\mathrm{cos2xdx}=\frac{\mathrm{sin2x}}{\mathrm{2}}+\mathrm{C} \\ $$
Commented by DAVONG last updated on 11/Jul/22
$$\mathrm{Thanks}\:\mathrm{sire} \\ $$