Question Number 219350 by deleted50 last updated on 23/Apr/25
$$\int\:\:\frac{\mathrm{1}}{\mathrm{cos}\left({u}\right)+\mathrm{sin}\left({u}\right)+\mathrm{1}}\:\mathrm{d}{u}=?? \\ $$
Answered by vnm last updated on 23/Apr/25
$$\int\frac{\mathrm{1}}{\mathrm{2cos}^{\mathrm{2}} \frac{{u}}{\mathrm{2}}+\mathrm{2sin}\frac{{u}}{\mathrm{2}}\mathrm{cos}\frac{{u}}{\mathrm{2}}}\mathrm{d}{u}= \\ $$$$\int\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{tan}\frac{{u}}{\mathrm{2}}\right)\centerdot\mathrm{2cos}^{\mathrm{2}} \frac{{u}}{\mathrm{2}}}\mathrm{d}{u}= \\ $$$$\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}\frac{{u}}{\mathrm{2}}}\mathrm{d}\left(\mathrm{1}+\mathrm{tan}\frac{{u}}{\mathrm{2}}\right)= \\ $$$$\mathrm{ln}\mid\mathrm{1}+\mathrm{tan}\frac{{u}}{\mathrm{2}}\mid+{C} \\ $$