Question Number 55996 by maxmathsup by imad last updated on 07/Mar/19
$${find}\:\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{x}}{{cosx}}{dx}\: \\ $$
Commented by maxmathsup by imad last updated on 09/Mar/19
![let I =∫_(π/3) ^(π/2) (x/(cosx))dx changement tan((x/2))=t give =∫_(1/(√3)) ^1 ((2arctan(t))/((1−t^2 )/(1+t^2 ))) ((2dt)/(1+t^2 )) =4∫_(1/(√3)) ^1 ((arctant)/(1−t^2 )) dt but ∫_(1/(√3)) ^1 ((arctan(t))/(1−t^2 ))dt =(1/2)∫_(1/(√3)) ^1 arctan(t){(1/(1−t)) +(1/(1+t))}dt =(1/2) ∫_(1/(√3)) ^1 ((arctan(t))/(1−t))dt +(1/2) ∫_(1/(√3)) ^1 ((arctan(t))/(1+t))dt let f(x) =∫_(1/(√3)) ^1 ((arctan(xt))/(1−t)) dt ⇒ f^′ (x) = ∫_(1/(√3)) ^1 (t/((1+x^2 t^2 )(1−t)))dt =_(xt =u) ∫_(x/(√3)) ^x (u/(x(1+u^2 )(1−(u/x)))) (du/x) =(1/x) ∫_(x/(√3)) ^x (u/((u^2 +1)(x−u))) du let decompose F(u) =(u/((x−u)(u^2 +1))) F(u) =(a/(x−u)) +((bu +c)/(u^2 +1)) a =lim_(u→x) (x−u)F(u) =(x/(x^2 +1)) lim_(u→+∞) uF(u) =0 =−a +b ⇒b=a ⇒F(u) =(a/(x−u)) +((au +c)/(u^2 +1)) F(0) =0 =(a/x) +c ⇒c =−(a/x) =−(1/(x^2 +1)) ⇒ F(u) =(x/((x^2 +1)(x−u))) +(((x/(x^2 +1))u −(1/(x^2 +1)))/(u^2 +1)) =(x/(x^2 +1)){(1/(x−u)) +((u−x)/(u^2 +1))} ⇒ ∫ F(u)du =(x/(x^2 +1))ln∣x−u∣ +(x/(2(x^2 +1)))ln(u^2 +1) −(x^2 /(x^2 +1)) arctan(u) +c ⇒ ∫_(x/(√3)) ^x F(u)du =(x/(x^2 +1))[ln∣x−u∣+(1/2)ln(u^2 +1)−x arctanu]_(x/(√3)) ^x ....be continued....](Q56062.png)
$${let}\:{I}\:=\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{x}}{{cosx}}{dx}\:\:{changement}\:\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:\:{give} \\ $$$$\:=\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{2}{arctan}\left({t}\right)}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\mathrm{4}\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:\:\:\frac{{arctant}}{\mathrm{1}−{t}^{\mathrm{2}} }\:{dt}\:\:{but} \\ $$$$\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({t}\right)}{\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:{arctan}\left({t}\right)\left\{\frac{\mathrm{1}}{\mathrm{1}−{t}}\:+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right\}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({t}\right)}{\mathrm{1}−{t}}{dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({t}\right)}{\mathrm{1}+{t}}{dt}\:\:\:{let}\:{f}\left({x}\right)\:=\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({xt}\right)}{\mathrm{1}−{t}}\:{dt}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:\frac{{t}}{\left(\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}\right)}{dt}\:\:=_{{xt}\:={u}} \:\:\:\:\int_{\frac{{x}}{\sqrt{\mathrm{3}}}} ^{{x}} \:\:\:\:\frac{{u}}{{x}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}−\frac{{u}}{{x}}\right)}\:\frac{{du}}{{x}} \\ $$$$=\frac{\mathrm{1}}{{x}}\:\int_{\frac{{x}}{\sqrt{\mathrm{3}}}} ^{{x}} \:\:\:\frac{{u}}{\left({u}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−{u}\right)}\:{du}\:\:{let}\:{decompose}\:{F}\left({u}\right)\:=\frac{{u}}{\left({x}−{u}\right)\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${F}\left({u}\right)\:=\frac{{a}}{{x}−{u}}\:+\frac{{bu}\:+{c}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{u}\rightarrow{x}} \left({x}−{u}\right){F}\left({u}\right)\:=\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${lim}_{{u}\rightarrow+\infty} {uF}\left({u}\right)\:=\mathrm{0}\:=−{a}\:+{b}\:\Rightarrow{b}={a}\:\Rightarrow{F}\left({u}\right)\:=\frac{{a}}{{x}−{u}}\:+\frac{{au}\:+{c}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}\:=\frac{{a}}{{x}}\:+{c}\:\Rightarrow{c}\:=−\frac{{a}}{{x}}\:=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−{u}\right)}\:+\frac{\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{u}\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\left\{\frac{\mathrm{1}}{{x}−{u}}\:+\frac{{u}−{x}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\right\}\:\Rightarrow\:\int\:{F}\left({u}\right){du}\:=\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{ln}\mid{x}−{u}\mid\:\:+\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}{ln}\left({u}^{\mathrm{2}} \:+\mathrm{1}\right) \\ $$$$−\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \:+\mathrm{1}}\:{arctan}\left({u}\right)\:+{c}\:\Rightarrow \\ $$$$\int_{\frac{{x}}{\sqrt{\mathrm{3}}}} ^{{x}} \:{F}\left({u}\right){du}\:=\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\left[{ln}\mid{x}−{u}\mid+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)−{x}\:{arctanu}\right]_{\frac{{x}}{\sqrt{\mathrm{3}}}} ^{{x}} \\ $$$$....{be}\:{continued}.... \\ $$