Question Number 41052 by turbo msup by abdo last updated on 01/Aug/18
$${find}\:{the}\:{value}\:{of}\:\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{x}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dxr} \\ $$
Commented by maxmathsup by imad last updated on 02/Aug/18
![let A = ∫_(π/6) ^(π/4) (x/(1+cos^2 x)) dx we have A =∫_(π/6) ^(π/4) (x/(1+(1/(1+tan^2 x))))dx = ∫_(π/6) ^(π/4) ((x(1+tan^2 x))/(2+tan^2 x)) dx changement tanx =t give A = ∫_(1/(√3)) ^1 ((arctant(1+t^2 ))/(2+t^2 )) (dt/(1+t^2 )) = ∫_(1/(√3)) ^1 ((arctant)/(2+t^2 )) dt let introduce the parametric function ϕ(x)= ∫_(1/(√3)) ^1 ((arctan(xt))/(2+t^2 )) dt we have ϕ^′ (x)= ∫_(1/(√3)) ^1 (t/((1+x^2 t^2 )(2+t^2 )))dt =_(xt =u) ∫_(x/(√3)) ^x (u/(x(1+u^2 )(2+(u^2 /x^2 )))) (du/x) = ∫_(x/(√3)) ^x (u/((1+u^2 )(2x^2 +u^2 ))) du let decompose F(u) = (u/((u^2 +1)(u^2 +2x^2 ))) F(u) = ((au+b)/(u^2 +1)) +((cu +d)/(u^2 +2x^2 )) F(−u) =−F(u) ⇒((−au+b)/(u^2 +1)) +((−cu +d)/(u^2 +2x^2 )) =((−au−b)/(u^2 +1)) +((−cu−d)/(u^2 +2x^2 )) ⇒ b=d=0 ⇒F(u) =((au)/(u^2 +1)) +((cu)/(u^2 +2x^2 )) lim_(u→+∞) uF(u) =0=a+c ⇒c=−a ⇒F(u)=((au)/(u^2 +1)) −((au)/(u^2 +2x^2 )) F(1) = (1/(2(1+2x^2 ))) =(a/2) −(a/(1+2x^2 )) ⇒(1/2) =((1+2x^2 )/2)a −a⇒ 1=(1+2x^2 )a−2a =(2x^2 −1)a ⇒a=(1/(2x^2 −1)) ⇒ F(u) =(1/(2x^2 −1)){ (u/(u^2 +1)) −(u/(u^2 +2x^2 ))}⇒ϕ^′ (x)=(1/(2x^2 −1)) ∫_(x/(√3)) ^x ((u/(u^2 +1)) −(u/(u^2 +2x^2 )))du = (1/(2(2x^2 −1)))[ ln∣((u^2 +1)/(u^2 +2x^2 ))∣]_(x/(√3)) ^x = (1/(2(2x^2 −1))){ln(((x^2 +1)/(3x^2 )))−ln((((x^2 /3)+1)/((x^2 /3) +2x^2 )))} = (1/(2(2x^2 −1))){ ln(((x^2 +1)/(3x^2 ))) −ln(((x^2 +3)/(7x^2 )))} =(1/(2(2x^2 −1))){ ln(x^2 +1)−ln(3)−2ln∣x∣ −ln(x^2 +3) +ln(7) +2ln∣x∣} =((ln(7)−ln(3))/(2(2x^2 −1))) +((ln(x^2 +1))/(2(2x^2 −1))) −((ln(x^2 +3))/(2(2x^2 −1))) ⇒ ϕ(x) = ∫ ((ln(7)−ln(3))/(2(2x^2 −1))) dx + ∫ ((ln(x^2 +1))/(2(2x^2 −1))) dx −∫ ((ln(x^2 +3))/(2(2x^2 −1))) dx +c ...be continued ....](Q41123.png)
$${let}\:\:{A}\:=\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\frac{{x}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}\:{dx}\:{we}\:{have}\:{A}\:=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{x}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}}{dx} \\ $$$$=\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{x}\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)}{\mathrm{2}+{tan}^{\mathrm{2}} {x}}\:{dx}\:\:{changement}\:\:{tanx}\:={t}\:{give} \\ $$$${A}\:=\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:\:\:\:\frac{{arctant}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{2}+{t}^{\mathrm{2}} }\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:\:\:\frac{{arctant}}{\mathrm{2}+{t}^{\mathrm{2}} }\:{dt}\:\:{let}\:{introduce}\:{the}\: \\ $$$${parametric}\:{function}\:\:\varphi\left({x}\right)=\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({xt}\right)}{\mathrm{2}+{t}^{\mathrm{2}} }\:{dt}\:\:{we}\:{have} \\ $$$$\varphi^{'} \left({x}\right)=\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:\:\:\:\frac{{t}}{\left(\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} \right)\left(\mathrm{2}+{t}^{\mathrm{2}} \right)}{dt}\:=_{{xt}\:={u}} \:\:\:\:\:\:\int_{\frac{{x}}{\sqrt{\mathrm{3}}}} ^{{x}} \:\:\:\:\:\frac{{u}}{{x}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{2}+\frac{{u}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)}\:\frac{{du}}{{x}} \\ $$$$=\:\int_{\frac{{x}}{\sqrt{\mathrm{3}}}} ^{{x}} \:\:\:\:\frac{{u}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{2}{x}^{\mathrm{2}} \:+{u}^{\mathrm{2}} \right)}\:{du}\:\:{let}\:{decompose}\:{F}\left({u}\right)\:=\:\frac{{u}}{\left({u}^{\mathrm{2}} +\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+\mathrm{2}{x}^{\mathrm{2}} \right)} \\ $$$${F}\left({u}\right)\:=\:\frac{{au}+{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:+\mathrm{2}{x}^{\mathrm{2}} } \\ $$$${F}\left(−{u}\right)\:=−{F}\left({u}\right)\:\Rightarrow\frac{−{au}+{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{cu}\:+{d}}{{u}^{\mathrm{2}} \:+\mathrm{2}{x}^{\mathrm{2}} }\:=\frac{−{au}−{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{cu}−{d}}{{u}^{\mathrm{2}} \:+\mathrm{2}{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${b}={d}=\mathrm{0}\:\Rightarrow{F}\left({u}\right)\:=\frac{{au}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{cu}}{{u}^{\mathrm{2}} \:+\mathrm{2}{x}^{\mathrm{2}} } \\ $$$${lim}_{{u}\rightarrow+\infty} {uF}\left({u}\right)\:=\mathrm{0}={a}+{c}\:\Rightarrow{c}=−{a}\:\Rightarrow{F}\left({u}\right)=\frac{{au}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{{au}}{{u}^{\mathrm{2}} \:+\mathrm{2}{x}^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right)}\:=\frac{{a}}{\mathrm{2}}\:−\frac{{a}}{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}{a}\:−{a}\Rightarrow \\ $$$$\mathrm{1}=\left(\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right){a}−\mathrm{2}{a}\:=\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right){a}\:\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}\left\{\:\frac{{u}}{{u}^{\mathrm{2}} +\mathrm{1}}\:−\frac{{u}}{{u}^{\mathrm{2}} \:+\mathrm{2}{x}^{\mathrm{2}} }\right\}\Rightarrow\varphi^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}\:\int_{\frac{{x}}{\sqrt{\mathrm{3}}}} ^{{x}} \:\left(\frac{{u}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{{u}}{{u}^{\mathrm{2}} \:+\mathrm{2}{x}^{\mathrm{2}} }\right){du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)}\left[\:{ln}\mid\frac{{u}^{\mathrm{2}} +\mathrm{1}}{{u}^{\mathrm{2}} \:+\mathrm{2}{x}^{\mathrm{2}} }\mid\right]_{\frac{{x}}{\sqrt{\mathrm{3}}}} ^{{x}} \:=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)}\left\{{ln}\left(\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}} }\right)−{ln}\left(\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{3}}+\mathrm{1}}{\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\:+\mathrm{2}{x}^{\mathrm{2}} }\right)\right\} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)}\left\{\:\:{ln}\left(\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}} }\right)\:−{ln}\left(\frac{{x}^{\mathrm{2}} \:+\mathrm{3}}{\mathrm{7}{x}^{\mathrm{2}} }\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)}\left\{\:\:{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−{ln}\left(\mathrm{3}\right)−\mathrm{2}{ln}\mid{x}\mid\:−{ln}\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)\:+{ln}\left(\mathrm{7}\right)\:+\mathrm{2}{ln}\mid{x}\mid\right\} \\ $$$$=\frac{{ln}\left(\mathrm{7}\right)−{ln}\left(\mathrm{3}\right)}{\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)}\:\:+\frac{{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)}\:−\frac{{ln}\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)}{\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)}\:\Rightarrow \\ $$$$\varphi\left({x}\right)\:=\:\int\:\:\frac{{ln}\left(\mathrm{7}\right)−{ln}\left(\mathrm{3}\right)}{\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)}\:{dx}\:+\:\int\:\:\:\:\:\frac{{ln}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)}\:{dx}\:−\int\:\:\frac{{ln}\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)}{\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)}\:{dx}\:+{c} \\ $$$$...{be}\:{continued}\:.... \\ $$