Question Number 68596 by Abdo msup. last updated on 14/Sep/19
$${calculate}\:\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\:\frac{{xdx}}{\mathrm{3}+{cosx}} \\ $$
Commented by mathmax by abdo last updated on 15/Sep/19
![changeent tan((x/2))=t give I =∫_(π/2) ^(π/3) ((xdx)/(3+cosx)) =∫_1 ^(1/( (√3))) ((2arctan(t))/(3+((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_1 ^(1/( (√3))) ((4arctan(t))/(3+3t^2 +1−t^2 ))dt =4 ∫_1 ^(1/( (√3))) ((arctan(t))/(2t^2 +4))dt =2 ∫_1 ^(1/( (√3))) ((arctan(t))/(t^2 +2))dt let f(α) =∫_1 ^(1/( (√3))) ((arctan(αt))/(t^(2 ) +2))dt with α>0 ⇒ f^′ (α)=∫_1 ^(1/( (√3))) (t/((α^2 t^2 +1)(t^2 +2)))dt let decompose F(t) =(t/((α^2 t^2 +1)(t^2 +2))) ⇒F(t)=((at+b)/(α^2 t^2 +1)) +((ct +d)/(t^2 +2)) F(−t)=−F(t) ⇒b=d=0 ⇒F(t) =((at)/(α^2 t^2 +1)) +((ct)/(t^2 +2)) lim_(t→+∞) tF(t) =0 =(a/α^2 ) +c ⇒c=−(a/α^2 ) ⇒ F(t) =((at)/(α^2 t^2 +1))−((at)/(α^2 (t^2 +2))) F(1) =(1/(3(α^2 +1))) =(a/(α^2 +1)) −(a/(3α^2 )) ⇒(1/3) =a−(a/(3α^2 ))(α^2 +1) ⇒ 1 =3a−((a(α^(2 ) +1))/α^2 ) =(((3α^2 −α^2 −1)a)/α^2 ) ⇒(2α^2 −1)a =α^2 ⇒ a =(α^2 /(2α^2 −1)) ⇒F(t) =((α^2 t)/((2α^2 −1)(α^2 t^2 +1))) −(t/((2α^2 −1)(t^2 +2))) ⇒ f^′ (α) =(1/(2(2α^2 −1))) ∫_1 ^(1/( (√3))) ((2α^2 t)/(α^2 t^2 +1))dt−(1/(2(2α^2 −1))) ∫_1 ^(1/( (√3))) ((2tdt)/(t^2 +2)) =(1/(2(2α^2 −1)))[ln(α^2 t^2 +1)]_1 ^(1/( (√3))) −(1/(2(2α^2 −1))) [ln(t^(2 ) +2)]_1 ^(1/( (√3))) =(1/(2(2α^2 −1))){ln((α^2 /3)+1)−ln(α^2 +1)}−(1/(2(2α^2 −1))){ln((1/3)+2)−ln(3)} ⇒f(α) =∫((ln(α^2 +3)−ln(α^2 +1)−ln(3))/(2(2α^2 −1)))dα −(ln((7/3))−ln(3))∫ (dα/(2(2α^2 −1))) +c....becontinued....](https://www.tinkutara.com/question/Q68694.png)
$${changeent}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:=\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{{xdx}}{\mathrm{3}+{cosx}}\:=\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\:\:\frac{\mathrm{2}{arctan}\left({t}\right)}{\mathrm{3}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\frac{\mathrm{4}{arctan}\left({t}\right)}{\mathrm{3}+\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:=\mathrm{4}\:\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\frac{{arctan}\left({t}\right)}{\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{4}}{dt}\:=\mathrm{2}\:\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\frac{{arctan}\left({t}\right)}{{t}^{\mathrm{2}} \:+\mathrm{2}}{dt} \\ $$$${let}\:{f}\left(\alpha\right)\:=\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\:\:\frac{{arctan}\left(\alpha{t}\right)}{{t}^{\mathrm{2}\:} +\mathrm{2}}{dt}\:\:\:{with}\:\alpha>\mathrm{0}\:\Rightarrow \\ $$$${f}^{'} \left(\alpha\right)=\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\:\frac{{t}}{\left(\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{2}\right)}{dt}\:\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\frac{{t}}{\left(\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{2}\right)}\:\Rightarrow{F}\left({t}\right)=\frac{{at}+{b}}{\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{2}} \\ $$$${F}\left(−{t}\right)=−{F}\left({t}\right)\:\Rightarrow{b}={d}=\mathrm{0}\:\Rightarrow{F}\left({t}\right)\:=\frac{{at}}{\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{ct}}{{t}^{\mathrm{2}} \:+\mathrm{2}} \\ $$$${lim}_{{t}\rightarrow+\infty} \:{tF}\left({t}\right)\:=\mathrm{0}\:=\frac{{a}}{\alpha^{\mathrm{2}} }\:+{c}\:\Rightarrow{c}=−\frac{{a}}{\alpha^{\mathrm{2}} }\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{{at}}{\alpha^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}}−\frac{{at}}{\alpha^{\mathrm{2}} \left({t}^{\mathrm{2}} \:+\mathrm{2}\right)} \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{3}\left(\alpha^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{{a}}{\alpha^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{{a}}{\mathrm{3}\alpha^{\mathrm{2}} }\:\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\:={a}−\frac{{a}}{\mathrm{3}\alpha^{\mathrm{2}} }\left(\alpha^{\mathrm{2}} \:+\mathrm{1}\right)\:\Rightarrow \\ $$$$\mathrm{1}\:=\mathrm{3}{a}−\frac{{a}\left(\alpha^{\mathrm{2}\:} +\mathrm{1}\right)}{\alpha^{\mathrm{2}} }\:=\frac{\left(\mathrm{3}\alpha^{\mathrm{2}} −\alpha^{\mathrm{2}} −\mathrm{1}\right){a}}{\alpha^{\mathrm{2}} }\:\Rightarrow\left(\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{1}\right){a}\:=\alpha^{\mathrm{2}} \:\Rightarrow \\ $$$${a}\:=\frac{\alpha^{\mathrm{2}} }{\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow{F}\left({t}\right)\:=\frac{\alpha^{\mathrm{2}} {t}}{\left(\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{1}\right)\left(\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:−\frac{{t}}{\left(\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{2}\right)}\:\Rightarrow \\ $$$${f}^{'} \left(\alpha\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{1}\right)}\:\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\:\frac{\mathrm{2}\alpha^{\mathrm{2}} {t}}{\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}}{dt}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{1}\right)}\:\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\:\frac{\mathrm{2}{tdt}}{{t}^{\mathrm{2}} \:+\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{1}\right)}\left[{ln}\left(\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{1}\right)}\:\left[{ln}\left({t}^{\mathrm{2}\:} +\mathrm{2}\right)\right]_{\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{1}\right)}\left\{{ln}\left(\frac{\alpha^{\mathrm{2}} }{\mathrm{3}}+\mathrm{1}\right)−{ln}\left(\alpha^{\mathrm{2}} \:+\mathrm{1}\right)\right\}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{1}\right)}\left\{{ln}\left(\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{2}\right)−{ln}\left(\mathrm{3}\right)\right\} \\ $$$$\Rightarrow{f}\left(\alpha\right)\:=\int\frac{{ln}\left(\alpha^{\mathrm{2}} +\mathrm{3}\right)−{ln}\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)−{ln}\left(\mathrm{3}\right)}{\mathrm{2}\left(\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{1}\right)}{d}\alpha \\ $$$$−\left({ln}\left(\frac{\mathrm{7}}{\mathrm{3}}\right)−{ln}\left(\mathrm{3}\right)\right)\int\:\:\frac{{d}\alpha}{\mathrm{2}\left(\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{1}\right)}\:+{c}….{becontinued}…. \\ $$$$ \\ $$