Question Number 200602 by Calculusboy last updated on 20/Nov/23
Answered by Frix last updated on 21/Nov/23
$$\mathrm{Use}\:{u}'=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:{t}}\:\rightarrow\:{u}=\mathrm{tan}\:\frac{{t}}{\mathrm{2}}; \\ $$$$\:\:\:\:\:\:\:\:\:{v}={t}+\mathrm{sin}\:{t}\:\rightarrow\:{v}'=\mathrm{1}+\mathrm{cos}\:{t} \\ $$$$\int\frac{{t}+\mathrm{sin}\:{t}}{\mathrm{1}+\mathrm{cos}\:{t}}{dt}= \\ $$$$=\left({t}+\mathrm{sin}\:{t}\right)\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:−\int\left(\mathrm{1}+\mathrm{cos}\:{t}\right)\:\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:{dt}= \\ $$$$={t}\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:+\mathrm{1}−\mathrm{cos}\:{t}\:−\int\mathrm{sin}\:{t}\:{dt}= \\ $$$$={t}\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:+{C} \\ $$
Commented by Calculusboy last updated on 21/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$