Question Number 36937 by maxmathsup by imad last updated on 07/Jun/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{x}\:{dx}}{\mathrm{1}+{cosx}} \\ $$
Commented by math khazana by abdo last updated on 08/Jun/18
![I =∫_0 ^(π/2) ((xdx)/(1+cosx)) + ∫_(π/2) ^π ((xdx)/(1+cosx)) but ∫_(π/2) ^π ((xdx)/(1+cosx)) =_(x=(π/2) +t) ∫_0 ^(π/2) ((((π/2)+t)dt)/(1−sint)) =(π/2) ∫_0 ^(π/2) (dt/(1−sint)) + ∫_0 ^(π/2) ((t sint)/(1−sin(t)))dt ∫_0 ^(π/2) ((xdx)/(1+cosx)) =_(tan((x/2))=t) ∫_0 ^1 ((2arctan(t))/(1+((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =4 ∫_0 ^1 ((arctan(t))/(1+t^2 +1−t^2 ))dt =2 ∫_0 ^1 arctant dt =2 { [tarctan(t)]_0 ^1 −∫_0 ^1 (t/(1+t^2 ))dt} =2{(π/4) −(1/2)[ln(1+t^2 )]_0 ^1 } =(π/2) −ln(2) ∫_0 ^(π/2) ((tsin(t))/(1−sin(t)))dt =_(tan((t/2))=x) ∫_0 ^1 ((2arctant ((2t)/(1+t^2 )))/(1−((2t)/(1+t^2 ))))dt = 4 ∫_0 ^1 ((t arctan(t))/(1+t^2 −2t)) dt =4 ∫_0 ^1 ((t arctan(t))/((t−1)^2 ))dt by parts ∫_0 ^1 ((t arctan(t))/((t−1)^2 ))dt =∫_0 ^1 (((t−1+1)arctan(t))/((t−1)^2 ))dt =∫_0 ^1 ((arctant)/(t−1))dt +∫_0 ^1 ((arctan(t))/((t−1)^2 ))dt but ∫_0 ^1 ((arctan(t))/(1−t))dt =∫_0 ^1 (Σ_(n=0) ^∞ t^n )arctant dt =Σ_(n=0) ^∞ ∫_0 ^1 t^n arctan(t) dt =Σ_(n=0) ^∞ A_n A_n = t^n arctant ∼ A t^n because arctan is borned ⇒ ∫_0 ^1 t^n arctant dt ∼ (A/(n+1)) and Σ A_n diverges so this integral diverges and we do the same manner for the other integral so I diverges..!](Q37070.png)
$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{xdx}}{\mathrm{1}+{cosx}}\:+\:\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:\:\:\frac{{xdx}}{\mathrm{1}+{cosx}}\:\:{but} \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:\:\frac{{xdx}}{\mathrm{1}+{cosx}}\:=_{{x}=\frac{\pi}{\mathrm{2}}\:+{t}} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\left(\frac{\pi}{\mathrm{2}}+{t}\right){dt}}{\mathrm{1}−{sint}} \\ $$$$=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{\mathrm{1}−{sint}}\:\:+\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{t}\:{sint}}{\mathrm{1}−{sin}\left({t}\right)}{dt} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{xdx}}{\mathrm{1}+{cosx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{arctan}\left({t}\right)}{\mathrm{1}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{arctan}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{arctant}\:{dt} \\ $$$$=\mathrm{2}\:\left\{\:\:\left[{tarctan}\left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\right\} \\ $$$$=\mathrm{2}\left\{\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \right\}\:=\frac{\pi}{\mathrm{2}}\:−{ln}\left(\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{tsin}\left({t}\right)}{\mathrm{1}−{sin}\left({t}\right)}{dt}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={x}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{arctant}\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{dt} \\ $$$$=\:\mathrm{4}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{t}\:{arctan}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} \:−\mathrm{2}{t}}\:{dt}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}\:{arctan}\left({t}\right)}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$${by}\:{parts}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}\:{arctan}\left({t}\right)}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left({t}−\mathrm{1}+\mathrm{1}\right){arctan}\left({t}\right)}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctant}}{{t}−\mathrm{1}}{dt}\:\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({t}\right)}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{arctan}\left({t}\right)}{\mathrm{1}−{t}}{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\sum_{{n}=\mathrm{0}} ^{\infty} {t}^{{n}} \right){arctant}\:{dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{{n}} \:{arctan}\left({t}\right)\:{dt}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \\ $$$${A}_{{n}} =\:{t}^{{n}} \:{arctant}\:\:\sim\:{A}\:{t}^{{n}} \:\:\:{because}\:{arctan}\:{is}\:{borned} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{n}} \:{arctant}\:{dt}\:\sim\:\frac{{A}}{{n}+\mathrm{1}}\:{and}\:\Sigma\:{A}_{{n}} \:{diverges} \\ $$$${so}\:{this}\:{integral}\:{diverges}\:{and}\:{we}\:{do}\:{the}\:{same} \\ $$$${manner}\:{for}\:{the}\:{other}\:{integral}\:{so}\:{I}\:{diverges}..! \\ $$$$ \\ $$$$ \\ $$