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1-1-cot-3x-dx-

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

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