Question Number 40716 by ajeetyadav4370 last updated on 26/Jul/18
$$\int\left({cosx}−{cos}\mathrm{2}{x}/\mathrm{1}−{cosx}\right){dx} \\ $$
Answered by maxmathsup by imad last updated on 26/Jul/18
$${let}\:{I}\:=\:\int\:\:\frac{{cosx}\:−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{1}−{cosx}}\:{dx} \\ $$$${I}\:=\:\int\:\:\:\frac{{cosx}\:−\left(\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\right)}{\mathrm{1}−{cosx}}{dx}\:=\:\int\:\:\frac{−\mathrm{2}{cos}^{\mathrm{2}} {x}\:+{cosx}\:+\mathrm{1}}{\mathrm{1}−{cosx}}{dx} \\ $$$$=\:\int\:\:\frac{\mathrm{1}−{cos}^{\mathrm{2}} {x}\:+{cosx}\:−{cos}^{\mathrm{2}} {x}}{\mathrm{1}−{cosx}}\:{dx} \\ $$$$=\:\int\:\:\frac{\left(\mathrm{1}−{cosx}\right)\left(\mathrm{1}+{cosx}\right)\:+{cosx}\left(\mathrm{1}−{cosx}\right)}{\mathrm{1}−{cosx}}{dx} \\ $$$$=\int\:\:\frac{\left(\mathrm{1}−{cosx}\right)\left(\mathrm{1}+\mathrm{2}{cosx}\right)}{\mathrm{1}−{cosx}}{dx}\:=\int\:\left(\mathrm{1}+\mathrm{2}{cosx}\right){dx}\:=\:{x}\:+\mathrm{2}{sinx}\:+{k} \\ $$$${I}\:={x}\:+\mathrm{2}{sinx}\:+{k}\:. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18
$$\int\frac{\mathrm{2}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}}{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx} \\ $$$$\int\frac{{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}}{{sin}\frac{{x}}{\mathrm{2}}}{dx} \\ $$$$\int\frac{\mathrm{3}{sin}\frac{{x}}{\mathrm{2}}−\mathrm{4}{sin}^{\mathrm{3}} \frac{{x}}{\mathrm{2}}}{{sin}\frac{{x}}{\mathrm{2}}}{dx} \\ $$$$\int\mathrm{3}−\mathrm{4}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\:{dx} \\ $$$$\mathrm{3}\int{dx}−\mathrm{2}\int\left(\mathrm{1}−{cosx}\right){dx} \\ $$$$\int{dx}+\mathrm{2}\int{cosx}\:{dx} \\ $$$${x}+\mathrm{2}{sinx}\:+{c} \\ $$