Question Number 166192 by greogoury55 last updated on 15/Feb/22
$$\:\:\:\:\int\:\frac{{dx}}{\mathrm{5}+\mathrm{4sin}\:{x}}\:=? \\ $$
Answered by som(math1967) last updated on 15/Feb/22
$$\:\:\int\frac{{dx}}{\mathrm{5}\left({sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)+\mathrm{8}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}} \\ $$$$=\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}}{\mathrm{5}{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{5}+\mathrm{8}{tan}\frac{{x}}{\mathrm{2}}} \\ $$$${let}\:{tan}\frac{{x}}{\mathrm{2}}={z}\Rightarrow{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}=\mathrm{2}{dz} \\ $$$$=\mathrm{2}\int\frac{{dz}}{\mathrm{5}{z}^{\mathrm{2}} +\mathrm{8}{z}+\mathrm{5}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\int\frac{{dz}}{{z}^{\mathrm{2}} +\frac{\mathrm{8}{z}}{\mathrm{5}}\:+\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\int\frac{{dz}}{{z}^{\mathrm{2}} +\mathrm{2}.{z}.\frac{\mathrm{4}}{\mathrm{5}}+\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{1}−\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\int\frac{{dz}}{\left({z}+\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}×\frac{\mathrm{5}}{\mathrm{3}}×\mathrm{tan}^{−\mathrm{1}} \left(\frac{{z}+\frac{\mathrm{4}}{\mathrm{5}}}{\frac{\mathrm{3}}{\mathrm{5}}}\right)\:+{C} \\ $$$${now}\:{put}\:{z}={tan}\frac{{x}}{\mathrm{2}} \\ $$
Commented by peter frank last updated on 15/Feb/22
$$\mathrm{thank}\:\mathrm{you} \\ $$