Question Number 45600 by maxmathsup by imad last updated on 14/Oct/18
$${find}\:{f}\left({x},{y}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({x}+{y}\:{sin}\theta\right){d}\theta\:\:{with}\:\:\mid{y}\mid<\mid{x}\mid \\ $$$$\left.\mathrm{2}\right)\:{find}\:{f}\left(\mathrm{2},\mathrm{3}\right) \\ $$$$\left.\mathrm{3}\right){find}\:{f}\left(\sqrt{\mathrm{2}},\sqrt{\mathrm{3}}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 15/Oct/18
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\frac{\partial{f}}{\partial{x}}\left({x},{y}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{{x}+{ysin}\theta}\:{changement}\:{tan}\left(\frac{\theta}{\mathrm{2}}\right)={t}\:{give} \\ $$$$\frac{\partial{f}}{\partial{x}}\left({x},{y}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({x}+{y}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right){x}\:+\mathrm{2}{yt}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{dt}}{{x}\:+{xt}^{\mathrm{2}} \:+\mathrm{2}{yt}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{dt}}{{xt}^{\mathrm{2}} \:+\mathrm{2}{yt}\:+{x}}\:\:{roots}\:{of}\:{p}\left({x}\right)={xt}^{\mathrm{2}} \:+\mathrm{2}{yt}\:+{x} \\ $$$$\Delta^{'} ={y}^{\mathrm{2}} −{x}^{\mathrm{2}} \:<\mathrm{0}\:\Rightarrow{no}\:{roots}\:{but}\:{xt}^{\mathrm{2}} \:+\mathrm{2}{yt}\:+{x}\:\:={x}\left({t}^{\mathrm{2}} \:+\mathrm{2}\frac{{y}}{{x}}{t}\:+\mathrm{1}\right) \\ $$$$={x}\left(\:\:\left({t}+\frac{{y}}{{x}}\right)^{\mathrm{2}} \:+\mathrm{1}−\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)\:={x}\left\{\:\left({t}+\frac{{y}}{{x}}\right)^{\mathrm{2}} \:+\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right\}\:{changement}\: \\ $$$${t}+\frac{{y}}{{x}}\:=\frac{\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}{\mid{x}\mid}\:{u}\:{give}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{dt}}{{xt}^{\mathrm{2}} \:+\mathrm{2}{yt}\:+{x}}\:=\:\int_{\frac{\mid{x}\mid}{{x}}\frac{{y}}{\:\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}} ^{\frac{\mid{x}\mid}{{x}}\frac{{x}+{y}}{\:\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}} \:\:\:\frac{\mathrm{2}}{{x}\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\left\{\mathrm{1}+{u}^{\mathrm{2}} \right\}}\frac{\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}{\mid{x}\mid}{du} \\ $$$$=\int_{\frac{{y}\xi\left({x}\right)}{\:\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}} ^{\frac{\left({x}+{y}\right)\xi\left({x}\right)}{\:\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}} \:\:\:\frac{\mathrm{2}}{\xi\left({x}\right)\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\xi\left({x}\right)\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\:\left\{\:{arctan}\left(\frac{\left({x}+{y}\right)\xi\left({x}\right)}{\:\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\right)−{arctan}\left(\frac{{y}\xi\left({x}\right)}{\:\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\right)\right\}=\frac{\partial{f}}{\partial{x}}\left({x},{y}\right)\:\Rightarrow \\ $$$${f}\left({x},{y}\right)\:=\frac{\mathrm{2}}{\xi\left({x}\right)}\:\int\:\:\:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\:{arctan}\left(\frac{\left({x}+{y}\right)\xi\left({x}\right)}{\:\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\right){dx}−\frac{\mathrm{2}}{\xi\left({x}\right)}\:\int\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}{arctan}\left(\frac{{y}\xi\left({x}\right)}{\:\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\right){dx}+{k} \\ $$$$\xi\left({x}\right)=\mathrm{1}\:{if}\:{x}>\mathrm{0}\:{and}\:\xi\left({x}\right)=−\mathrm{1}\:{if}\:{x}<\mathrm{0}\:…{be}\:{continued}… \\ $$
Commented by maxmathsup by imad last updated on 15/Oct/18
![2)f(2,3)=∫_0 ^(π/2) ln(2+3sinθ)dθ f(2,3)= ∫_0 ^(π/2) ln(3((2/3)+sinθ))dθ =(π/2)ln(3) +∫_0 ^(π/2) ln((2/3)+sinθ)dθ let find ϕ(x)=∫_0 ^(π/2) ln(x+sinθ)dθ with ∣x∣<1 we have ϕ^′ (x) = ∫_0 ^(π/2) (dθ/(x+sinθ)) =_(tan((θ/2))=t) ∫_0 ^1 (1/(x+((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^1 ((2dt)/(x(1+t^2 ) +2t)) = ∫_0 ^1 ((2dt)/(x +xt^2 +2t)) = ∫_0 ^1 ((2dt)/(xt^2 +2t +x)) Δ^′ =1−x^2 >0 ⇒t_1 =((−1+(√(1−x^2 )))/x) and t_2 =((−1−(√(1−x^2 )))/x) (we take x≠0) F(t)=(1/(xt^2 +2t +x)) =(1/(x(t−t_1 )(t−t_2 ))) =(1/x)(x/(2(√(1−x^2 )))){ (1/(t−t_1 )) −(1/(t−t_2 ))} =(1/(2(√(1−x^2 )))){ (1/(t−t_1 )) −(1/(t−t_2 ))} ⇒∫_0 ^1 F(t)dt=(1/(2(√(1−x^2 ))))[ln∣((t−t_1 )/(t−t_2 ))∣]_0 ^1 =(1/(2(√(1−x^2 )))){ ln∣((1−t_1 )/(1−t_2 ))∣−ln∣(t_1 /t_2 )∣}=(1/(2(√(1−x^2 ))))ln∣((1−t_1 )/(1−t_2 )).(t_2 /t_1 )∣ =(1/(2(√(1−x^2 ))))ln∣((t_2 −t_1 t_2 )/(t_1 −t_1 t_2 ))∣ =(1/(2(√(1−x^2 ))))ln∣ ((((−1−(√(1−x^2 )))/x) −1)/(((−1+(√(1−x^2 )))/x)−1))∣ =(1/(2(√(1−x^2 ))))ln∣ ((1+((1+(√(1−x^2 )))/x))/(1−((−1+(√(1−x^2 )))/x)))∣=(1/(2(√(1−x^2 ))))ln∣ ((x+1+(√(1−x^2 )))/(x+1−(√(1−x^2 ))))∣ ⇒ ϕ^′ (x) =(1/( (√(1−x^2 ))))ln(((x+1+(√(1−x^2 )))/(x+1−(√(1−x^2 ))))) ⇒ ϕ(x) =∫_0 ^x (1/( (√(1−u^2 ))))ln(((u+1+(√(1−u^2 )))/(u+1−(√(1−u^2 )))))du +c c=ϕ(0)=∫_0 ^(π/2) ln(sinθ)dθ =−(π/2)ln(2) ⇒ ϕ(x) =∫_0 ^x (1/( (√(1−u^2 ))))ln(((u+1+(√(1−u^2 )))/(u+1−(√(1−u^2 )))))du−(π/2)ln(2) ...be continued...](https://www.tinkutara.com/question/Q45713.png)
$$\left.\mathrm{2}\right){f}\left(\mathrm{2},\mathrm{3}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{2}+\mathrm{3}{sin}\theta\right){d}\theta\:\: \\ $$$${f}\left(\mathrm{2},\mathrm{3}\right)=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{3}\left(\frac{\mathrm{2}}{\mathrm{3}}+{sin}\theta\right)\right){d}\theta\:=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{3}\right)\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{2}}{\mathrm{3}}+{sin}\theta\right){d}\theta \\ $$$${let}\:{find}\:\varphi\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({x}+{sin}\theta\right){d}\theta\:\:{with}\:\mid{x}\mid<\mathrm{1}\:\:{we}\:{have} \\ $$$$\varphi^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{{x}+{sin}\theta}\:=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={t}} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{{x}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{dt}}{{x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:+\mathrm{2}{t}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{dt}}{{x}\:+{xt}^{\mathrm{2}} \:+\mathrm{2}{t}}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{dt}}{{xt}^{\mathrm{2}} \:+\mathrm{2}{t}\:+{x}} \\ $$$$\Delta^{'} =\mathrm{1}−{x}^{\mathrm{2}} >\mathrm{0}\:\:\Rightarrow{t}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:\:{and}\:{t}_{\mathrm{2}} =\frac{−\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:\:\left({we}\:{take}\:{x}\neq\mathrm{0}\right) \\ $$$${F}\left({t}\right)=\frac{\mathrm{1}}{{xt}^{\mathrm{2}} \:+\mathrm{2}{t}\:+{x}}\:=\frac{\mathrm{1}}{{x}\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{{x}}\frac{{x}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left\{\:\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left\{\:\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right\}\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {F}\left({t}\right){dt}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left[{ln}\mid\frac{{t}−{t}_{\mathrm{1}} }{{t}−{t}_{\mathrm{2}} }\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left\{\:{ln}\mid\frac{\mathrm{1}−{t}_{\mathrm{1}} }{\mathrm{1}−{t}_{\mathrm{2}} }\mid−{ln}\mid\frac{{t}_{\mathrm{1}} }{{t}_{\mathrm{2}} }\mid\right\}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{ln}\mid\frac{\mathrm{1}−{t}_{\mathrm{1}} }{\mathrm{1}−{t}_{\mathrm{2}} }.\frac{{t}_{\mathrm{2}} }{{t}_{\mathrm{1}} }\mid \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{ln}\mid\frac{{t}_{\mathrm{2}} −{t}_{\mathrm{1}} {t}_{\mathrm{2}} }{{t}_{\mathrm{1}} −{t}_{\mathrm{1}} {t}_{\mathrm{2}} }\mid\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{ln}\mid\:\frac{\frac{−\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:−\mathrm{1}}{\frac{−\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}−\mathrm{1}}\mid \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{ln}\mid\:\frac{\mathrm{1}+\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}}{\mathrm{1}−\frac{−\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}}\mid=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{ln}\mid\:\frac{{x}+\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}+\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\mid\:\Rightarrow \\ $$$$\varphi^{'} \left({x}\right)\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{ln}\left(\frac{{x}+\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}+\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)\:\Rightarrow \\ $$$$\varphi\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{ln}\left(\frac{{u}+\mathrm{1}+\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{{u}+\mathrm{1}−\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}\right){du}\:+{c} \\ $$$${c}=\varphi\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\theta\right){d}\theta\:=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\varphi\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} \:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{ln}\left(\frac{{u}+\mathrm{1}+\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{{u}+\mathrm{1}−\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}\right){du}−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:…{be}\:{continued}… \\ $$