Question Number 204961 by naka3546 last updated on 04/Mar/24
$$\int\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cot}\:\mathrm{3}{x}}\:{dx}\:=\:\: \\ $$
Answered by TonyCWX08 last updated on 04/Mar/24
$${let}\:{t}\:=\:\mathrm{3}{x} \\ $$$${dt}\:=\:\mathrm{3}\:{dx} \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\int\frac{\mathrm{1}}{\mathrm{1}+{cot}\left(\mathrm{3}{x}\right)}×\mathrm{3}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{1}}{\mathrm{1}+{cot}\left({t}\right)}\:{dt} \\ $$$$ \\ $$$${Using}\:{Weierstrass}\:{Substitution} \\ $$$${cot}\left({t}\right)=\frac{\mathrm{1}−{tan}\left(\frac{{t}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}{tan}\left(\frac{{t}}{\mathrm{2}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}−{tan}\left(\frac{{t}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}{tan}\left(\frac{{t}}{\mathrm{2}}\right)}}\:{dt} \\ $$$$ \\ $$$${let}\:{u}\:=\:{tan}\left(\frac{{t}}{\mathrm{2}}\right) \\ $$$${du}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{tan}\left(\frac{{t}}{\mathrm{2}}\right)^{\mathrm{2}} \right) \\ $$$$ \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{2}{u}}}×\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{1}}{\frac{\mathrm{2}{u}−\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{2}{u}}}×\frac{\mathrm{1}}{\frac{\mathrm{1}+{u}^{\mathrm{2}} }{\mathrm{2}}}\:{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\:\int\:\frac{\mathrm{2}{u}}{\mathrm{2}{u}−\mathrm{1}−{u}^{\mathrm{2}} }×\frac{\mathrm{2}}{\mathrm{1}+{u}^{\mathrm{2}} }\:{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\:\int\:\frac{\mathrm{4}{u}}{\left(\mathrm{2}{u}−\mathrm{1}−{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:{du} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{3}}\int\:\left(\frac{{u}−\mathrm{1}}{\mathrm{4}\left(\mathrm{2}{u}−\mathrm{1}−{u}^{\mathrm{2}} \right)}+\frac{{u}+\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\right)\:{du} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{3}}\left(\int\:\frac{{u}−\mathrm{1}}{\mathrm{4}\left(\mathrm{2}{u}−\mathrm{1}−{u}^{\mathrm{2}} \right)}\:{du}\:+\:\int\:\frac{{u}+\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:{du}\:\right) \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{3}}\left(−\frac{\mathrm{1}}{\mathrm{8}}\left({ln}\left(\mid\mathrm{2}{u}+\mathrm{1}−{u}^{\mathrm{2}} \mid\right)\right)+\frac{\mathrm{1}}{\mathrm{8}}\left({ln}\left(\mid\mathrm{1}+{u}^{\mathrm{2}} \mid\right)\right)+\frac{\mathrm{tan}^{−\mathrm{1}} \left({u}\right)}{\mathrm{4}}\right) \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{3}}\:\left(−\frac{\mathrm{1}}{\mathrm{8}}\left({ln}\left(\mathrm{2}{tan}\frac{\mathrm{3}{x}}{\mathrm{2}}+\mathrm{1}−{tan}\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)^{\mathrm{2}} \right)\right)+\frac{\mathrm{1}}{\mathrm{8}}\left({ln}\left(\mathrm{1}+{tan}\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)^{\mathrm{2}} \right)+\frac{{x}}{\mathrm{2}}\right.\right. \\ $$$$=−\frac{\mathrm{1}}{\mathrm{6}}\left({ln}\left(\mathrm{2}{tan}\left(\frac{\mathrm{2}{x}}{\mathrm{3}}\right)−\mathrm{1}−{tan}\left(\frac{\mathrm{2}{x}}{\mathrm{3}}\right)^{\mathrm{2}} \right)\right)+\frac{\mathrm{1}}{\mathrm{6}}\left({ln}\left(\mathrm{1}+{tan}\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)^{\mathrm{2}} \right)+\frac{{x}}{\mathrm{2}}+{C}\right. \\ $$
Commented by naka3546 last updated on 04/Mar/24
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}. \\ $$
Commented by Frix last updated on 04/Mar/24
$$\mathrm{From}\:\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dt}}{\mathrm{1}+\mathrm{cot}\:{t}}\:\mathrm{it}'\mathrm{s}\:\mathrm{enough}\:\mathrm{to}\:\mathrm{substitute} \\ $$$${u}=\mathrm{tan}\:{t}\:\rightarrow\:{dt}=\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}};\:\mathrm{cot}\:{t}\:=\frac{\mathrm{1}}{{u}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{u}}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{du}\:\mathrm{which}\:\mathrm{is}\:\mathrm{easier}\:\mathrm{to}\:\mathrm{solve}. \\ $$
Answered by Frix last updated on 04/Mar/24
$$\int\frac{{dx}}{\mathrm{1}+\mathrm{cot}\:\mathrm{3}{x}}\:\overset{{t}=\mathrm{tan}\:\mathrm{3}{x}} {=}\:\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\int\left(\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}}+\mathrm{tan}^{−\mathrm{1}} \:{t}\:−\mathrm{ln}\:\left({t}+\mathrm{1}\right)\right)= \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \:{t}\:+\frac{\mathrm{ln}\:\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }}{\mathrm{12}}= \\ $$$$=\frac{{x}}{\mathrm{2}}−\frac{\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{3}{x}\:+\mathrm{sin}\:\mathrm{3}{x}\mid}{\mathrm{6}}+{C} \\ $$