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Question-200602




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

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