cos-2-ln-tan-x-2-tan-x-2-dx-

Question Number 169268 by amin96 last updated on 27/Apr/22

$$\boldsymbol{\Omega}=\int\frac{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)\right)}{\boldsymbol{\mathrm{tan}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)}\boldsymbol{\mathrm{dx}}=? \\ $$

Answered by Mathspace last updated on 28/Apr/22

ln(tan((x/2)))=t ⇒tan((x/2))=e^t ⇒x=2arctan(e^t ) ⇒ Ψ=∫((cos^2 (t))/e^t )×((2e^t )/(1+e^(2t) ))dt =2∫ ((cos^2 (t))/(1+e^(2t) ))dt =∫ ((1+cos(2t))/(1+e^(2t) ))dt =∫ ((e^(−2t) (1+cos(2t)))/(1+e^(−2t) ))dt =∫ e^(−2t) Σ_(n=0) ^∞ (−1)^n e^(−2nt) dt +∫ e^(−2t) cos(2t)Σ_(n=0) ^∞ (−1)^n e^(−2nt) =Σ_(n=0) ^∞ (−1)^n ∫ e^(−2(n+1)t) dt +Σ_(n=0) ^∞ (−1)^n ∫ e^(−2(n+1)t) cos(2t)dt =I +J I=−(1/2)Σ_(n=0) ^∞ (((−1)^n )/(n+1))e^(−2(n+1)t) +c_1 J=Σ (−1)^n u_n u_n =Re(∫ e^(−2(n+1)t+2it) dt)=....

$${ln}\left({tan}\left(\frac{{x}}{\mathrm{2}}\right)\right)={t}\:\Rightarrow{tan}\left(\frac{{x}}{\mathrm{2}}\right)={e}^{{t}} \\ $$$$\Rightarrow{x}=\mathrm{2}{arctan}\left({e}^{{t}} \right)\:\Rightarrow \\ $$$$\Psi=\int\frac{{cos}^{\mathrm{2}} \left({t}\right)}{{e}^{{t}} }×\frac{\mathrm{2}{e}^{{t}} }{\mathrm{1}+{e}^{\mathrm{2}{t}} }{dt} \\ $$$$=\mathrm{2}\int\:\:\frac{{cos}^{\mathrm{2}} \left({t}\right)}{\mathrm{1}+{e}^{\mathrm{2}{t}} }{dt} \\ $$$$=\int\:\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{1}+{e}^{\mathrm{2}{t}} }{dt} \\ $$$$=\int\:\:\frac{{e}^{−\mathrm{2}{t}} \left(\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\right)}{\mathrm{1}+{e}^{−\mathrm{2}{t}} }{dt} \\ $$$$=\int\:\:{e}^{−\mathrm{2}{t}} \sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{e}^{−\mathrm{2}{nt}} {dt} \\ $$$$+\int\:{e}^{−\mathrm{2}{t}} {cos}\left(\mathrm{2}{t}\right)\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {e}^{−\mathrm{2}{nt}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\int\:\:{e}^{−\mathrm{2}\left({n}+\mathrm{1}\right){t}} {dt} \\ $$$$+\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \int\:{e}^{−\mathrm{2}\left({n}+\mathrm{1}\right){t}} {cos}\left(\mathrm{2}{t}\right){dt} \\ $$$$={I}\:+{J} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{e}^{−\mathrm{2}\left({n}+\mathrm{1}\right){t}} \:+{c}_{\mathrm{1}} \\ $$$${J}=\Sigma\:\left(−\mathrm{1}\right)^{{n}} {u}_{{n}} \\ $$$${u}_{{n}} ={Re}\left(\int\:\:{e}^{−\mathrm{2}\left({n}+\mathrm{1}\right){t}+\mathrm{2}{it}} {dt}\right)=…. \\ $$$$ \\ $$

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