Question Number 55493 by Kunal12588 last updated on 25/Feb/19
$${Find}\:{intgral}\:{of}\:\mathrm{tan}\:{x}\:{by}\:{parts} \\ $$$$\int{tan}\:{x}\:{dx}\:\:\:{pls}\:{help}\:{me}\:\:\left({i}\:{am}\:{a}\:{newbie}\:{in}\:{calculus}\right) \\ $$
Commented by Kunal12588 last updated on 25/Feb/19
$${i}\:{am}\:{getting}\:{something}\:{like}\:{this} \\ $$$$\int\left(\frac{\mathrm{1}}{{cosx}}\right)×\left({sin}\:{x}\:{dx}\right) \\ $$$${taking}\:{u}=\frac{\mathrm{1}}{{cosx}} \\ $$$$\Rightarrow{du}=\:{tan}\:{x}\:{sec}\:{x}\:{dx} \\ $$$${taking}\:{dv}={sinx}\:{dx} \\ $$$$\Rightarrow{v}=−{cosx} \\ $$$$\int{u}\:{dv}\:=\:{uv}\:−\:\int{v}\:{du} \\ $$$$\int{tan}\:{x}\:{dx}\:=\:\frac{\mathrm{1}}{{cos}\:{x}}×\left(−{cos}\:{x}\right)\:−\int−{cos}\:{x}\:×\:{tan}\:{x}\:{sec}\:{x}\:{dx} \\ $$$$\int{tan}\:{x}\:{dx}\:=\:−\mathrm{1}+\int{tan}\:{x}\:{dx} \\ $$$${but}\:{in}\:{the}\:{book}\::− \\ $$$$\int{tan}\:{x}\:{dx}\:=\:−{ln}\:{cos}\:{x} \\ $$
Commented by maxmathsup by imad last updated on 25/Feb/19
$${let}\:{I}\:=\int\:{tanx}\:{dx}\:\Rightarrow{I}\:=\int\:\frac{{sinx}}{{cosx}}{dx}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\int\:\:\frac{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\frac{\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{dt}\:=\int\:\:\mathrm{2}{t}\left\{\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }\right\}{dt} \\ $$$$=\int\:\:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}\:−\int\:\:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}\:\:={ln}\mid{t}^{\mathrm{2}} +\mathrm{1}\mid−{ln}\mid{t}^{\mathrm{2}} −\mathrm{1}\mid\:+{c} \\ $$$$={ln}\mid\frac{\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}\mid\:+{c}\:={ln}\mid\frac{\mathrm{1}+\frac{{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}}{\mathrm{1}−\frac{{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}}\mid\:+{c} \\ $$$$={ln}\mid\:\frac{\mathrm{1}}{{cos}\left({x}\right)}\mid\:+{c}\:={ln}\mid{cosx}\mid\:+{C}\:. \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 25/Feb/19
$${what}\:{is}\:{the}\:{question}\:\int{tanxdx}\rightarrow{by}\:{parts} \\ $$$${or}\:{by}\:\boldsymbol{{partial}}\:\boldsymbol{{fraction}}…\boldsymbol{{pls}}\:\boldsymbol{{recheck}}\:\boldsymbol{{the}}\:\boldsymbol{{question}}.. \\ $$
Commented by maxmathsup by imad last updated on 25/Feb/19
$${I}\:=−{ln}\mid{cosx}\mid\:+{C}\:\:. \\ $$
Commented by maxmathsup by imad last updated on 25/Feb/19
$$\:{i}\:{have}\:{used}\:{changement}\:{method}\:{not}\:{integration}\:{by}\:{parts}… \\ $$
Commented by Kunal12588 last updated on 26/Feb/19
$${sir}\:{it}\:{was}\:{not}\:{a}\:{question}\:{I}\:{was}\:{trying}\:{by}\:{my} \\ $$$${own}.\:{I}\:{want}\:{to}\:{know}\:{where}\:{is}\:{the}\:{mistake} \\ $$