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Question Number 40044 by abdo mathsup 649 cc last updated on 15/Jul/18

let f(t) = ∫_0 ^(π/2) ln( cosx +t sinx) 1) calculate f(0) 2) calculate f^′ (t) then find a simple form of f(t) 3) calculate ∫_0 ^(π/2) ln(cosx +2 sinx)dx 4) calculate ∫_0 ^(π/2) ln((√3)cosx +sinx)dx

$${let}\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\:{cosx}\:+{t}\:{sinx}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{'} \left({t}\right)\:{then}\:{find}\:\:{a}\:{simple}\:{form}\:{of}\:{f}\left({t}\right) \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cosx}\:+\mathrm{2}\:{sinx}\right){dx} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\sqrt{\mathrm{3}}{cosx}\:+{sinx}\right){dx} \\ $$

Commented by abdo mathsup 649 cc last updated on 15/Jul/18

f(t) =∫_0 ^(π/2) ln(cosx +t sinx)dx.

$${f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cosx}\:+{t}\:{sinx}\right){dx}. \\ $$

Commented by maxmathsup by imad last updated on 22/Jul/18

1) we have f(t)= ∫_0 ^(π/2) ln(cosx+t sinx)dx ⇒f(0)=∫_0 ^(π/2) ln(cosx)dx=−(π/2)ln(2) (this value is proved ) 2)f^′ (t) =∫_0 ^(π/2) (∂/∂t){ln(cosx +tsinx)}dx =∫_0 ^(π/2) ((sinx)/(cosx+tsinx))dx hangement tan((x/2)) =u give f^′ (t) = ∫_0 ^1 (((2u)/(1+u^2 ))/(((1−u^2 )/(1+u^2 )) +t ((2u)/(1+u^2 )))) ((2du)/(1+u^2 )) = ∫_0 ^1 ((4u du)/((1+u^2 ){1−u^2 +2tu})) = −∫_0 ^1 ((4udu)/((1+u^2 ){u^2 −2tu−1} )) let decompose F(u) = ((4u)/((u^2 +1)(u^2 −2tu −1))) roots of u^2 −2tu −1 Δ^′ = t^2 +1 ⇒u_1 = t+(√(1+t^2 )) and u_2 =t−(√(1+t^2 )) and F(u)= ((4u)/((u−u_1 )(u−u_2 )(1+u^2 ))) F(u) = (a/(u−u_1 )) +(b/(u−u_2 )) + ((cu +d)/(u^2 +1)) a =lim_(u→u_1 ) (u−u_1 )F(u) = ((4u_1 )/((u_1 −u_2 )(1+u_1 ^2 ))) = ((4(t+(√(1+t^2 ))))/(2(√(1+t^2 ))( 1+(t+(√(1+t^2 )))^2 )) b =lim_(u→u_2 ) (u−u_2 )F(u) =((4u_2 )/((u_2 −u_1 )(1+u_2 ^2 ))) =((4(t−(√(1+t^2 ))))/(−2(√(1+t^2 )){1+(t−(√(1+t^2 )))^2 )) lim_(u→+∞) u F(u) =0 =a+b +c ⇒c =−a−b ⇒ F(u) = (a/(u−u_1 )) +(b/(u−u_2 )) +(((−a−b)u +d)/(u^2 +1)) F(0) =0 =−(a/u_1 ) −(b/u_2 ) +d ⇒d = (a/u_1 ) +(b/u_2 ) ....be continued...

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({t}\right)=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cosx}+{t}\:{sinx}\right){dx}\:\Rightarrow{f}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cosx}\right){dx}=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\ $$$$\left({this}\:{value}\:{is}\:{proved}\:\right) \\ $$$$\left.\mathrm{2}\right){f}^{'} \left({t}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\partial}{\partial{t}}\left\{{ln}\left({cosx}\:+{tsinx}\right)\right\}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sinx}}{{cosx}+{tsinx}}{dx}\:\:{hangement}\:\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:={u}\:{give} \\ $$$${f}^{'} \left({t}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\:+{t}\:\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{4}{u}\:{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left\{\mathrm{1}−{u}^{\mathrm{2}} +\mathrm{2}{tu}\right\}} \\ $$$$=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{4}{udu}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left\{{u}^{\mathrm{2}} −\mathrm{2}{tu}−\mathrm{1}\right\}\:}\:\:\:{let}\:{decompose}\:{F}\left({u}\right)\:=\:\:\:\frac{\mathrm{4}{u}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}^{\mathrm{2}} −\mathrm{2}{tu}\:−\mathrm{1}\right)} \\ $$$${roots}\:{of}\:\:{u}^{\mathrm{2}} \:−\mathrm{2}{tu}\:−\mathrm{1} \\ $$$$\Delta^{'} \:=\:{t}^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow{u}_{\mathrm{1}} =\:{t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\:{and}\:\:{u}_{\mathrm{2}} ={t}−\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\:\:\:\:{and}\:{F}\left({u}\right)=\:\frac{\mathrm{4}{u}}{\left({u}−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$${F}\left({u}\right)\:=\:\frac{{a}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}−{u}_{\mathrm{2}} }\:+\:\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{u}\rightarrow{u}_{\mathrm{1}} } \:\:\left({u}−{u}_{\mathrm{1}} \right){F}\left({u}\right)\:=\:\:\:\frac{\mathrm{4}{u}_{\mathrm{1}} }{\left({u}_{\mathrm{1}} −{u}_{\mathrm{2}} \right)\left(\mathrm{1}+{u}_{\mathrm{1}} ^{\mathrm{2}} \right)}\:=\:\frac{\mathrm{4}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)}{\mathrm{2}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\left(\:\mathrm{1}+\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} \right.} \\ $$$${b}\:={lim}_{{u}\rightarrow{u}_{\mathrm{2}} } \:\:\:\left({u}−{u}_{\mathrm{2}} \right){F}\left({u}\right)\:=\frac{\mathrm{4}{u}_{\mathrm{2}} }{\left({u}_{\mathrm{2}} −{u}_{\mathrm{1}} \right)\left(\mathrm{1}+{u}_{\mathrm{2}} ^{\mathrm{2}} \right)}\:=\frac{\mathrm{4}\left({t}−\sqrt{\left.\mathrm{1}+{t}^{\mathrm{2}} \right)}\right.}{−\mathrm{2}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\left\{\mathrm{1}+\left({t}−\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} \right.} \\ $$$${lim}_{{u}\rightarrow+\infty} {u}\:{F}\left({u}\right)\:=\mathrm{0}\:={a}+{b}\:+{c}\:\Rightarrow{c}\:=−{a}−{b}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\:\frac{{a}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}−{u}_{\mathrm{2}} }\:+\frac{\left(−{a}−{b}\right){u}\:+{d}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}\:=−\frac{{a}}{{u}_{\mathrm{1}} }\:−\frac{{b}}{{u}_{\mathrm{2}} }\:\:+{d}\:\:\Rightarrow{d}\:=\:\frac{{a}}{{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}_{\mathrm{2}} }\:\:\:....{be}\:{continued}... \\ $$

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