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cosx-2-cosx-dx-

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Question Number 27614 by NECx last updated on 10/Jan/18
∫((cosx)/(2−cosx))dx
$$\int\frac{{cosx}}{\mathrm{2}−{cosx}}{dx} \\ $$
Commented by abdo imad last updated on 11/Jan/18
∫((cosx)/(2−cosx))dx= −∫((2−cosx +2)/(2−cosx))dx = −x +2∫ (dx/(2−cox)) but the ch. tan((x/2))=t give ∫(dx/(2−cosx))= ∫ (((2dt)/(1+t^2 ))/(2−((1−t^2 )/(1+t^2 ))))= ∫ ((2dt)/(2+2t^2 −1+t^2 )) = ∫ ((2dt)/(3t^2 +1)) = (2/3) ∫ (dt/(t^2 + (((√3)/3))^2 ))=_(t= ((√3)/3) u) (2/3) ∫((((√3)/3)du)/((((√3)/3))^2 (1+u^2 ))) =(2/3).(3/( (√3))) arctanu +λ= (2/( (√3))) arctan((√3) tan((x/2))) ∫((cosx)/(2−cosx))dx= −x + (4/( (√3))) arctan((√3) tan((x/2))) +λ
$$\int\frac{{cosx}}{\mathrm{2}−{cosx}}{dx}=\:−\int\frac{\mathrm{2}−{cosx}\:+\mathrm{2}}{\mathrm{2}−{cosx}}{dx}\:=\:−{x}\:+\mathrm{2}\int\:\frac{{dx}}{\mathrm{2}−{cox}}\:\:{but} \\ $$$${the}\:{ch}.\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give}\: \\ $$$$\int\frac{{dx}}{\mathrm{2}−{cosx}}=\:\int\:\frac{\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{2}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}=\:\:\int\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} \:−\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\:\int\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{1}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\int\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{2}} }=_{{t}=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:{u}} \:\:\frac{\mathrm{2}}{\mathrm{3}}\:\int\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{du}}{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{2}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}}}\:{arctanu}\:+\lambda=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\sqrt{\mathrm{3}}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$\int\frac{{cosx}}{\mathrm{2}−{cosx}}{dx}=\:−{x}\:\:+\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\sqrt{\mathrm{3}}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\right)\:+\lambda \\ $$
Commented by NECx last updated on 11/Jan/18
yeah.... Thanks... You′ve cleared my doubts.
$${yeah}….\:{Thanks}…\:{You}'{ve}\:{cleared} \\ $$$${my}\:{doubts}. \\ $$

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