Question Number 55267 by maxmathsup by imad last updated on 25/Feb/19
![1) calculate f(x)=∫_0 ^(π/4) ln(1+xtanθ)dθ 2) find the values of integrals ∫_0 ^(π/4) ln(1+tanθ) and ∫_0 ^(π/4) ln(1+2tanθ)dθ . 1) we have f^′ (x)=∫_0 ^(π/4) ((tanθ)/(1+xtanθ)) dθ =∫_0 ^(π/4) (((sinθ)/(cosθ))/(1+x((sinθ)/(cosθ))))dθ =∫_0 ^(π/4) ((sinθ)/(cosθ +xsinθ)) dθ =_(tan((θ/2))=t) ∫_0 ^((√2)−1) (((2t)/(1+t^2 ))/(((1−t^2 )/(1+t^2 )) +((2xt)/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^((√2)−1) ((4t)/((1+t^2 )(1−t^2 +2xt)))dt =−∫_0 ^((√2)−1) ((4t)/((t^2 +1)(t^2 −2xt −1)))dt let decompose F(t) = ((4t)/((t^2 +1)(t^2 −2xt −1))) roots of t^2 −2xt −1 Δ^′ =x^2 +1 ⇒t_1 =x+(√(x^2 +1)) and t_2 =x−(√(x^2 +1)) F(t)=(a/(t−t_1 )) +(b/(t−t_2 )) +((ct +d)/(t^2 +1)) a =lim_(t→t_1 ) (t−t_1 )F(t)=((4t_1 )/((t_1 ^2 +1)(t_1 −t_2 ))) =α b =lim_(t→t_2 ) (t−t_2 )F(t) =((4t_2 )/((t_2 ^2 +1)(t_2 −t_1 ))) =β ⇒F(t)=(α/(t−t_1 )) +(β/(t−t_2 )) +((ct +d)/(t^2 +1)) F(0) =0=−(α/t_1 ) −(β/t_2 ) +d ⇒d =(α/t_1 ) +(β/t_2 ) F(1)=(2/(−2x)) =−(1/x)=(α/(1−t_1 )) +(β/(1−t_2 )) +((c+d)/2) ⇒(1/x) =(α/(t_1 −1)) +(β/(t_2 −1)) −(c/2) −(d/2) ⇒(c/2) =(α/(t_1 −1)) +(β/(t_2 −1)) −(d/2) −(1/x) ⇒c =((2α)/(t_1 −1)) +((2β)/(t_2 −1)) −d−(2/x) ∫ F(t)dt =αln∣t−t_1 ∣ +βln∣t−t_2 ∣ +(c/2)ln(t^2 +1) +d arctan(t) ⇒ ∫_0 ^((√2)−1) F(t)dt =[αln∣t−t_1 ∣+βln∣t−t_2 ∣ +(c/2)ln(t^2 +1)]_0 ^((√2)−1) =αln∣(√2)−1−t_1 ∣ +βln∣(√2)−1−t_2 ∣ +(c/2)ln(4−2(√2)) =αln∣(√2)−1−x−(√(1+x^2 )))+βln∣(√2)−1−x+(√(1+x^2 ))) +((ln(4−2(√2)))/2)c =f^′ (x) ⇒ f(x)=∫ αln∣(√2)−1−x−(√(1+x^2 ))∣)dx+β∫ ln∣(√2)−1+(√(1+x^2 ))∣dx +((cx)/2)ln(4−2(√2)) +C ....be continued...](https://www.tinkutara.com/question/Q55267.png)
$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left(\mathrm{1}+{xtan}\theta\right){d}\theta \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{values}\:{of}\:{integrals}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left(\mathrm{1}+{tan}\theta\right)\:\:{and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+\mathrm{2}{tan}\theta\right){d}\theta\:. \\ $$$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{tan}\theta}{\mathrm{1}+{xtan}\theta}\:{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{\frac{{sin}\theta}{{cos}\theta}}{\mathrm{1}+{x}\frac{{sin}\theta}{{cos}\theta}}{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{sin}\theta}{{cos}\theta\:+{xsin}\theta}\:{d}\theta\:\:=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={t}} \:\:\:\:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\frac{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{2}{xt}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\:\frac{\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}^{\mathrm{2}} \:+\mathrm{2}{xt}\right)}{dt}\:=−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:−\mathrm{1}\right)}{dt}\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:−\mathrm{1}\right)}\:\:{roots}\:{of}\:\:{t}^{\mathrm{2}} −\mathrm{2}{xt}\:−\mathrm{1} \\ $$$$\Delta^{'} ={x}^{\mathrm{2}} +\mathrm{1}\:\Rightarrow{t}_{\mathrm{1}} ={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:{and}\:{t}_{\mathrm{2}} ={x}−\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left({t}\right)=\frac{{a}}{{t}−{t}_{\mathrm{1}} }\:+\frac{{b}}{{t}−{t}_{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{t}\rightarrow{t}_{\mathrm{1}} } \left({t}−{t}_{\mathrm{1}} \right){F}\left({t}\right)=\frac{\mathrm{4}{t}_{\mathrm{1}} }{\left({t}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{1}\right)\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)}\:=\alpha \\ $$$${b}\:={lim}_{{t}\rightarrow{t}_{\mathrm{2}} } \left({t}−{t}_{\mathrm{2}} \right){F}\left({t}\right)\:=\frac{\mathrm{4}{t}_{\mathrm{2}} }{\left({t}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)}\:=\beta\:\Rightarrow{F}\left({t}\right)=\frac{\alpha}{{t}−{t}_{\mathrm{1}} }\:+\frac{\beta}{{t}−{t}_{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}=−\frac{\alpha}{{t}_{\mathrm{1}} }\:−\frac{\beta}{{t}_{\mathrm{2}} }\:+{d}\:\:\Rightarrow{d}\:=\frac{\alpha}{{t}_{\mathrm{1}} }\:+\frac{\beta}{{t}_{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)=\frac{\mathrm{2}}{−\mathrm{2}{x}}\:=−\frac{\mathrm{1}}{{x}}=\frac{\alpha}{\mathrm{1}−{t}_{\mathrm{1}} }\:+\frac{\beta}{\mathrm{1}−{t}_{\mathrm{2}} }\:+\frac{{c}+{d}}{\mathrm{2}}\:\Rightarrow\frac{\mathrm{1}}{{x}}\:=\frac{\alpha}{{t}_{\mathrm{1}} −\mathrm{1}}\:+\frac{\beta}{{t}_{\mathrm{2}} −\mathrm{1}}\:−\frac{{c}}{\mathrm{2}}\:−\frac{{d}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{c}}{\mathrm{2}}\:=\frac{\alpha}{{t}_{\mathrm{1}} −\mathrm{1}}\:+\frac{\beta}{{t}_{\mathrm{2}} −\mathrm{1}}\:−\frac{{d}}{\mathrm{2}}\:−\frac{\mathrm{1}}{{x}}\:\Rightarrow{c}\:=\frac{\mathrm{2}\alpha}{{t}_{\mathrm{1}} −\mathrm{1}}\:+\frac{\mathrm{2}\beta}{{t}_{\mathrm{2}} −\mathrm{1}}\:−{d}−\frac{\mathrm{2}}{{x}} \\ $$$$\int\:{F}\left({t}\right){dt}\:=\alpha{ln}\mid{t}−{t}_{\mathrm{1}} \mid\:+\beta{ln}\mid{t}−{t}_{\mathrm{2}} \mid\:+\frac{{c}}{\mathrm{2}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\:+{d}\:{arctan}\left({t}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} {F}\left({t}\right){dt}\:=\left[\alpha{ln}\mid{t}−{t}_{\mathrm{1}} \mid+\beta{ln}\mid{t}−{t}_{\mathrm{2}} \mid\:+\frac{{c}}{\mathrm{2}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$$=\alpha{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{t}_{\mathrm{1}} \mid\:+\beta{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{t}_{\mathrm{2}} \mid\:+\frac{{c}}{\mathrm{2}}{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)\: \\ $$$$\left.=\left.\alpha{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\beta{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:+\frac{{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{2}}{c}\:={f}^{'} \left({x}\right)\:\Rightarrow \\ $$$$\left.{f}\left({x}\right)=\int\:\alpha{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\mid\right){dx}+\beta\int\:{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\mid{dx} \\ $$$$+\frac{{cx}}{\mathrm{2}}{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:+{C}\:….{be}\:{continued}… \\ $$