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cosx-2-cosx-dx-

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Question Number 77778 by Dah Solu Tion last updated on 10/Jan/20
∫((cosx)/(2−cosx))dx
$$\int\frac{{cosx}}{\mathrm{2}−{cosx}}{dx} \\ $$$$ \\ $$
Commented by Tony Lin last updated on 10/Jan/20
let t=tan(x/2), (dx/dt)=(2/(1+t^2 )) ∫(((1−t^2 )/(1+t^2 ))/(((2+2t^2 )/(1+t^2 ))−((1−t^2 )/(1+t^2 ))))×(2/(1+t^2 ))dt =2∫((1−t^2 )/(1+3t^2 ))×(1/(1+t^2 ))dt =2∫(2/(1+3t^2 ))−(1/(1+t^2 ))dt =4∫(1/(1+((√3)t)^2 ))dt−2∫(1/(1+t^2 ))dt =(4/( (√3)))tan^(−1) (√3)t−2tan^(−1) t+c =(4/( (√3)))tan^(−1) ((√3)tan(x/2))−x+c
$${let}\:{t}={tan}\frac{{x}}{\mathrm{2}},\:\frac{{dx}}{{dt}}=\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\int\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}×\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{2}\int\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{3}{t}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} \:}{dt} \\ $$$$=\mathrm{2}\int\frac{\mathrm{2}}{\mathrm{1}+\mathrm{3}{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{4}\int\frac{\mathrm{1}}{\mathrm{1}+\left(\sqrt{\mathrm{3}}{t}\right)^{\mathrm{2}} }{dt}−\mathrm{2}\int\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \sqrt{\mathrm{3}}{t}−\mathrm{2}{tan}^{−\mathrm{1}} {t}+{c} \\ $$$$=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{3}}{tan}\frac{{x}}{\mathrm{2}}\right)−{x}+{c} \\ $$
Commented by john santu last updated on 10/Jan/20

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