ln-tanx-dx-

Question Number 183123 by SANOGO last updated on 21/Dec/22

$$\int{ln}\left({tanx}\right){dx} \\ $$

Answered by TheSupreme last updated on 21/Dec/22

∫ln(((sin(x))/(cos(x))))dx t=tan(x) (dt/(1+t^2 )) =dx ∫((ln(t))/(1+t^2 ))dt=ln(t)arctan(t)−∫(1/(t(1+t^2 )))dt= ln(t)arctan(t)−∫(A/t)+((Bt+C)/(1+t^2 ))dt A+At^2 +Bt^2 +Ct=1 A=1 C=0 B=−1 I=ln(t)arctan(t)−∫(1/t)dt+∫(t/(1+t^2 ))dt I=ln(t)arctan(t)−ln(t)+(1/2)ln(1+t^2 ) I=ln(tan(x))(arctan(tan(x))−1)−ln(cos(x))

$$\int{ln}\left(\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)}\right){dx} \\ $$$${t}={tan}\left({x}\right) \\ $$$$\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:={dx} \\ $$$$\int\frac{{ln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}={ln}\left({t}\right){arctan}\left({t}\right)−\int\frac{\mathrm{1}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}= \\ $$$${ln}\left({t}\right){arctan}\left({t}\right)−\int\frac{{A}}{{t}}+\frac{{Bt}+{C}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$${A}+{At}^{\mathrm{2}} +{Bt}^{\mathrm{2}} +{Ct}=\mathrm{1} \\ $$$${A}=\mathrm{1} \\ $$$${C}=\mathrm{0} \\ $$$${B}=−\mathrm{1} \\ $$$${I}={ln}\left({t}\right){arctan}\left({t}\right)−\int\frac{\mathrm{1}}{{t}}{dt}+\int\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$${I}={ln}\left({t}\right){arctan}\left({t}\right)−{ln}\left({t}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right) \\ $$$${I}={ln}\left({tan}\left({x}\right)\right)\left({arctan}\left({tan}\left({x}\right)\right)−\mathrm{1}\right)−{ln}\left({cos}\left({x}\right)\right) \\ $$

Commented by MJS_new last updated on 21/Dec/22

wrong. why does noone test their results anymore? ∫((ln t)/(t^2 +1))dt= u′=(1/(t^2 +1)) → u=arctan t v=ln t → v′=(1/t) ∫u′v=uv−∫uv′ =ln t arctan t −∫((arctan t)/t)dt

$$\mathrm{wrong}.\:\mathrm{why}\:\mathrm{does}\:\mathrm{noone}\:\mathrm{test}\:\mathrm{their}\:\mathrm{results}\:\mathrm{anymore}? \\ $$$$\int\frac{\mathrm{ln}\:{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:{u}'=\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{u}=\mathrm{arctan}\:{t} \\ $$$$\:\:\:\:\:{v}=\mathrm{ln}\:{t}\:\rightarrow\:{v}'=\frac{\mathrm{1}}{{t}} \\ $$$$\:\:\:\:\:\int{u}'{v}={uv}−\int{uv}' \\ $$$$=\mathrm{ln}\:{t}\:\mathrm{arctan}\:{t}\:−\int\frac{\mathrm{arctan}\:{t}}{{t}}{dt} \\ $$

Commented by SANOGO last updated on 21/Dec/22

$${merci}\:{bien} \\ $$

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