Question Number 46853 by maxmathsup by imad last updated on 01/Nov/18
$${fnd}\:\:\int\:\:\:\:\:\:\frac{{dx}}{\mathrm{1}+{cos}\left({tx}\right)} \\ $$
Commented by maxmathsup by imad last updated on 01/Nov/18
$${let}\:{A}\left({t}\right)\:=\int\:\:\frac{{dx}}{\mathrm{1}+{cos}\left({tx}\right)}\:\Rightarrow{A}\left({t}\right)\:=_{{tx}={u}} \:\:\:\int\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{cosu}}\:\frac{{du}}{{t}} \\ $$$$=\frac{\mathrm{1}}{{t}}\:\int\:\:\:\:\frac{{du}}{\mathrm{1}+{cosu}}\:=_{{tan}\left(\frac{{u}}{\mathrm{2}}\right)=\alpha} \:\:\frac{\mathrm{1}}{{t}}\:\int\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}−\alpha^{\mathrm{2}} }{\mathrm{1}+\alpha^{\mathrm{2}} }}\:\frac{\mathrm{2}{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{{t}}\:\int\:\:\frac{\mathrm{2}{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} \:+\mathrm{1}−\alpha^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{t}}\:\int\:{d}\alpha\:=\frac{\alpha}{{t}}\:+{c}\:=\frac{{tan}\left(\frac{{u}}{\mathrm{2}}\right)}{{t}}\:+{c} \\ $$$$=\frac{{tan}\left(\frac{{tx}}{\mathrm{2}}\right)}{{t}}\:+{c}\:\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18
$$\int\frac{\mathrm{1}−{cos}\left({tx}\right)}{{sin}^{\mathrm{2}} \left({tx}\right)}{dx} \\ $$$$\int{cosec}^{\mathrm{2}} \left({tx}\right)−{cot}\left({tx}\right){cosec}\left({tx}\right)\:\:\:{dx} \\ $$$$\frac{−{cot}\left({tx}\right)}{{t}}+\frac{{cosec}\left({tx}\right)}{{t}}+{c} \\ $$