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Question Number 91910 by Ar Brandon last updated on 03/May/20
∫((ln(1+sin^2 x))/(sin^2 x))dx
$$\int\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$
Commented by mathmax by abdo last updated on 03/May/20
let f(a) =∫((ln(1+asin^2 x))/(sin^2 x)) with a>0 f^′ (a) =∫ln(1+asin^2 x)dx =∫ln(1+a×((1−cos(2x))/2))dx =∫ ln(2+a−acos(2x))dx−xln(2) =∫ln((2+a)(1−(a/(2+a))cos(2x))dx−xln(2) =(ln(2+a)−ln2)x +∫ ln(1−(a/(a+2))cos(2x)) let determine I_λ =∫ln(1−λcos(2x))dx with 0<λ<1 (dI_λ /dλ)=−∫ ((cos(2x))/(1−λcos(2x)))dx =_(2x=t) (1/2) ∫((cos(t))/(1−λcost))dt =−(1/(2λ))∫ ((1−λcost −1)/(1−λcost))dt =−(1/(2λ)) t +(1/(2λ)) ∫ (dt/(1−λcost)) ∫ (dt/(1−λcost)) =_(tan((x/2))=u) ∫ ((2du)/((1+u^2 )(1−λ((1−u^2 )/(1+u^2 ))))) =2 ∫ (du/(1+u^2 −λ +λu^2 )) =2 ∫ (du/((1+λ)u^2 +1−λ)) =(2/(1+λ)) ∫ (du/(u^2 +((1−λ)/(1+λ)))) =_(u=(√((1−λ)/(1+λ)))z) (2/(1+λ)) ∫ (1/(((1−λ)/(1+λ))(1+z^2 )))×((√(1−λ))/( (√(1+λ))))dz =(2/( (√(1−λ^2 ))))×(π/2) =(π/( (√(1−λ^2 )))) ⇒I ′_λ =−(x/λ) +(π/(2λ(√(1−λ^2 )))) ⇒ I_λ =−xlnλ +(π/2) ∫ (dλ/(λ(√(1−λ^2 )))) +C ∫ (dλ/(λ(√(1−λ^2 )))) =_(λ=sint) ∫ ((cost dt)/(sint cost)) =∫ (dt/(sint)) =_(tan((t/2))=u) =∫ ((2du)/((1+u^2 )((2u)/(1+u^2 )))) =∫ (du/u) =ln∣u∣ =ln∣tan(((arcsinλ)/2))∣ ⇒ I_λ =−xlnλ +(π/2)ln∣tan(((arcsinλ)/2))∣ +C ⇒ f^′ (a) =xln(((2+a)/2))−xln((a/(a+2)))+(π/2)ln∣tan(((arcsin((a/(a+2))))/2))∣ +C ⇒f(a) =x ∫(ln(((2+a)/2))−ln((a/(a+2)))∣+(π/2)∫ ln∣tan(((arcsin((a/(a+2))))/2)) +C ....be continued....
$${let}\:{f}\left({a}\right)\:=\int\frac{{ln}\left(\mathrm{1}+{asin}^{\mathrm{2}} {x}\right)}{{sin}^{\mathrm{2}} {x}}\:\:{with}\:{a}>\mathrm{0} \\ $$$${f}^{'} \left({a}\right)\:=\int{ln}\left(\mathrm{1}+{asin}^{\mathrm{2}} {x}\right){dx}\:=\int{ln}\left(\mathrm{1}+{a}×\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right){dx} \\ $$$$=\int\:{ln}\left(\mathrm{2}+{a}−{acos}\left(\mathrm{2}{x}\right)\right){dx}−{xln}\left(\mathrm{2}\right) \\ $$$$=\int{ln}\left(\left(\mathrm{2}+{a}\right)\left(\mathrm{1}−\frac{{a}}{\mathrm{2}+{a}}{cos}\left(\mathrm{2}{x}\right)\right){dx}−{xln}\left(\mathrm{2}\right)\right. \\ $$$$=\left({ln}\left(\mathrm{2}+{a}\right)−{ln}\mathrm{2}\right){x}\:+\int\:{ln}\left(\mathrm{1}−\frac{{a}}{{a}+\mathrm{2}}{cos}\left(\mathrm{2}{x}\right)\right)\:{let}\:{determine} \\ $$$${I}_{\lambda} =\int{ln}\left(\mathrm{1}−\lambda{cos}\left(\mathrm{2}{x}\right)\right){dx}\:\:{with}\:\mathrm{0}<\lambda<\mathrm{1} \\ $$$$\frac{{dI}_{\lambda} }{{d}\lambda}=−\int\:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\mathrm{1}−\lambda{cos}\left(\mathrm{2}{x}\right)}{dx}\:=_{\mathrm{2}{x}={t}} \frac{\mathrm{1}}{\mathrm{2}}\:\:\:\int\frac{{cos}\left({t}\right)}{\mathrm{1}−\lambda{cost}}{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\lambda}\int\:\:\frac{\mathrm{1}−\lambda{cost}\:−\mathrm{1}}{\mathrm{1}−\lambda{cost}}{dt}\:=−\frac{\mathrm{1}}{\mathrm{2}\lambda}\:{t}\:+\frac{\mathrm{1}}{\mathrm{2}\lambda}\:\int\:\:\frac{{dt}}{\mathrm{1}−\lambda{cost}} \\ $$$$\int\:\frac{{dt}}{\mathrm{1}−\lambda{cost}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={u}} \:\:\:\int\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}−\lambda\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)} \\ $$$$=\mathrm{2}\:\int\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} −\lambda\:+\lambda{u}^{\mathrm{2}} }\:=\mathrm{2}\:\int\:\:\frac{{du}}{\left(\mathrm{1}+\lambda\right){u}^{\mathrm{2}} \:+\mathrm{1}−\lambda} \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+\lambda}\:\int\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda}}\:=_{{u}=\sqrt{\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda}}{z}} \:\:\:\:\frac{\mathrm{2}}{\mathrm{1}+\lambda}\:\int\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda}\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}×\frac{\sqrt{\mathrm{1}−\lambda}}{\:\sqrt{\mathrm{1}+\lambda}}{dz} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }}×\frac{\pi}{\mathrm{2}}\:=\frac{\pi}{\:\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }}\:\Rightarrow{I}\:'_{\lambda} \:\:=−\frac{{x}}{\lambda}\:+\frac{\pi}{\mathrm{2}\lambda\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }}\:\Rightarrow \\ $$$${I}_{\lambda} =−{xln}\lambda\:\:+\frac{\pi}{\mathrm{2}}\:\int\:\:\frac{{d}\lambda}{\lambda\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }}\:+{C} \\ $$$$\int\:\frac{{d}\lambda}{\lambda\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }}\:=_{\lambda={sint}} \:\:\:\int\:\:\frac{{cost}\:{dt}}{{sint}\:{cost}}\:=\int\:\frac{{dt}}{{sint}}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \\ $$$$=\int\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:=\int\:\frac{{du}}{{u}}\:={ln}\mid{u}\mid\:={ln}\mid{tan}\left(\frac{{arcsin}\lambda}{\mathrm{2}}\right)\mid\:\Rightarrow \\ $$$${I}_{\lambda} =−{xln}\lambda\:+\frac{\pi}{\mathrm{2}}{ln}\mid{tan}\left(\frac{{arcsin}\lambda}{\mathrm{2}}\right)\mid\:+{C}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)\:={xln}\left(\frac{\mathrm{2}+{a}}{\mathrm{2}}\right)−{xln}\left(\frac{{a}}{{a}+\mathrm{2}}\right)+\frac{\pi}{\mathrm{2}}{ln}\mid{tan}\left(\frac{{arcsin}\left(\frac{{a}}{{a}+\mathrm{2}}\right)}{\mathrm{2}}\right)\mid\:+{C} \\ $$$$\Rightarrow{f}\left({a}\right)\:={x}\:\int\left({ln}\left(\frac{\mathrm{2}+{a}}{\mathrm{2}}\right)−{ln}\left(\frac{{a}}{{a}+\mathrm{2}}\right)\mid+\frac{\pi}{\mathrm{2}}\int\:{ln}\mid{tan}\left(\frac{{arcsin}\left(\frac{{a}}{{a}+\mathrm{2}}\right)}{\mathrm{2}}\right)\:+{C}\right. \\ $$$$….{be}\:{continued}…. \\ $$
Commented by Ar Brandon last updated on 03/May/20
�� thanks
Commented by mathmax by abdo last updated on 03/May/20
you are welcome
$${you}\:{are}\:{welcome} \\ $$
Commented by Ar Brandon last updated on 03/May/20
I got this. What do think ? ����
Commented by turbo msup by abdo last updated on 03/May/20
this is the way the probleme here is thst integrsl is without limits...
$${this}\:{is}\:{the}\:{way}\:{the}\:{probleme}\:{here} \\ $$$${is}\:{thst}\:{integrsl}\:{is}\:{without}\:{limits}… \\ $$
Answered by Ar Brandon last updated on 03/May/20

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