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Question Number 122406 by mathdave last updated on 16/Nov/20
solve ∫_0 ^(π/4) ln(tan^2 x+2)dx
$${solve} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{2}\right){dx} \\ $$
Answered by TANMAY PANACEA last updated on 16/Nov/20
f(x)=ln(2+tan^2 x) f(0)=ln2 f((π/4))=ln3 ∫_0 ^(π/4) ln3 dx>∫_0 ^(π/4) f(x)dx>∫_0 ^(π/4) lm2 dx (π/4)ln3>I>(π/4)ln2
$${f}\left({x}\right)={ln}\left(\mathrm{2}+{tan}^{\mathrm{2}} {x}\right) \\ $$$${f}\left(\mathrm{0}\right)={ln}\mathrm{2} \\ $$$${f}\left(\frac{\pi}{\mathrm{4}}\right)={ln}\mathrm{3} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\mathrm{3}\:{dx}>\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {f}\left({x}\right){dx}>\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {lm}\mathrm{2}\:{dx} \\ $$$$\frac{\pi}{\mathrm{4}}{ln}\mathrm{3}>{I}>\frac{\pi}{\mathrm{4}}{ln}\mathrm{2} \\ $$$$ \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 16/Nov/20
I =∫_0 ^(π/4) ln(1+tan^2 x+1) =∫_0 ^(π/4) ln(1+(1/(cos^2 x)))dx =∫_0 ^(π/4) ln(1+cos^2 x)dx−2∫_0 ^(π/4) ln(cosx)dx ∫_0 ^(π/4) ln(1+cos^2 x)dx =∫_0 ^(π/4) ln(1+((1+cos(2x))/2))dx =∫_0 ^(π/4) ln(3+cos(2x))dx−(π/4)ln(2) =(π/4)ln(3) +∫_0 ^(π/4) ln(1+(1/3)cos(2x))dx−(π/4)ln(2) ∫_0 ^(π/4) ln(1+(1/3)cos(2x))dx=_(2x=t) (1/2) ∫_0 ^(π/2) ln(1+(1/3)cost)dt let f(a) =∫_0 ^(π/2) ln(1+acost)dt with 0<a<1 f^′ (a) =∫_0 ^(π/2) ((cost)/(1+acost))dt =(1/a)∫_0 ^(π/2) ((1+acost−1)/(1+acost))dt =(π/(2a))−(1/a) ∫_0 ^(π/2) (dt/(1+acost))(→tan((t/2))=u) =(π/(2a))−(1/a) ∫_0 ^1 ((2du)/((1+u^2 )(1+a((1−u^2 )/(1+u^2 )))))=(π/(2a))−(2/a)∫_0 ^1 (du/(1+u^2 +a−au^2 )) =(π/(2a))−(2/a) ∫_0 ^1 (du/(1+a+(1−a)u^2 )) we have (2/a)∫_0 ^1 (du/(1+a +(1−a)u^2 )) =(2/(a(1+a)))∫_0 ^1 (du/(1+((1−a)/(1+a))u^2 )) =_((√((1−a)/(1+a)))u=z) (2/(a(1+a)))∫_0 ^(√((1−a)/(1+a))) (1/(1+z^2 ))×((√(1+a))/( (√(1−a))))dz =(2/(a(√(1−a^2 )))) arctan((√((1−a)/(1+a)))) ⇒ f(a) =(π/2)lna−2∫ ((arctan((√((1−a)/(1+a)))))/(a(√(1−a^2 ))))da +c ∫ ((arctan((√((1−a)/(1+a)))))/(a(√(1−a^2 ))))da =_(a=cosθ) ∫ ((arctan(tan((θ/2))))/(cosθ .sinθ)) sinθ dθ =(1/2)∫ (θ/(cosθ))dθ.....be continued....
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}+\mathrm{1}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)\mathrm{dx}−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cosx}\right)\mathrm{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)\mathrm{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\mathrm{ln}\left(\mathrm{3}+\mathrm{cos}\left(\mathrm{2x}\right)\right)\mathrm{dx}−\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$=\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{3}\right)\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}\left(\mathrm{2x}\right)\right)\mathrm{dx}−\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}\left(\mathrm{2x}\right)\right)\mathrm{dx}=_{\mathrm{2x}=\mathrm{t}} \:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cost}\right)\mathrm{dt}\:\mathrm{let} \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{acost}\right)\mathrm{dt}\:\:\mathrm{with}\:\mathrm{0}<\mathrm{a}<\mathrm{1} \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cost}}{\mathrm{1}+\mathrm{acost}}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{a}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+\mathrm{acost}−\mathrm{1}}{\mathrm{1}+\mathrm{acost}}\mathrm{dt} \\ $$$$=\frac{\pi}{\mathrm{2a}}−\frac{\mathrm{1}}{\mathrm{a}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{acost}}\left(\rightarrow\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)=\mathrm{u}\right) \\ $$$$=\frac{\pi}{\mathrm{2a}}−\frac{\mathrm{1}}{\mathrm{a}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2du}}{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{a}\frac{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\right)}=\frac{\pi}{\mathrm{2a}}−\frac{\mathrm{2}}{\mathrm{a}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{du}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} +\mathrm{a}−\mathrm{au}^{\mathrm{2}} } \\ $$$$=\frac{\pi}{\mathrm{2a}}−\frac{\mathrm{2}}{\mathrm{a}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{du}}{\mathrm{1}+\mathrm{a}+\left(\mathrm{1}−\mathrm{a}\right)\mathrm{u}^{\mathrm{2}} }\:\mathrm{we}\:\mathrm{have} \\ $$$$\frac{\mathrm{2}}{\mathrm{a}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{du}}{\mathrm{1}+\mathrm{a}\:+\left(\mathrm{1}−\mathrm{a}\right)\mathrm{u}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\mathrm{a}\left(\mathrm{1}+\mathrm{a}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{du}}{\mathrm{1}+\frac{\mathrm{1}−\mathrm{a}}{\mathrm{1}+\mathrm{a}}\mathrm{u}^{\mathrm{2}} } \\ $$$$=_{\sqrt{\frac{\mathrm{1}−\mathrm{a}}{\mathrm{1}+\mathrm{a}}}\mathrm{u}=\mathrm{z}} \:\:\:\:\frac{\mathrm{2}}{\mathrm{a}\left(\mathrm{1}+\mathrm{a}\right)}\int_{\mathrm{0}} ^{\sqrt{\frac{\mathrm{1}−\mathrm{a}}{\mathrm{1}+\mathrm{a}}}} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }×\frac{\sqrt{\mathrm{1}+\mathrm{a}}}{\:\sqrt{\mathrm{1}−\mathrm{a}}}\mathrm{dz} \\ $$$$=\frac{\mathrm{2}}{\mathrm{a}\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\:\mathrm{arctan}\left(\sqrt{\frac{\mathrm{1}−\mathrm{a}}{\mathrm{1}+\mathrm{a}}}\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=\frac{\pi}{\mathrm{2}}\mathrm{lna}−\mathrm{2}\int\:\:\frac{\mathrm{arctan}\left(\sqrt{\frac{\mathrm{1}−\mathrm{a}}{\mathrm{1}+\mathrm{a}}}\right)}{\mathrm{a}\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\mathrm{da}\:+\mathrm{c} \\ $$$$\int\:\:\:\frac{\mathrm{arctan}\left(\sqrt{\frac{\mathrm{1}−\mathrm{a}}{\mathrm{1}+\mathrm{a}}}\right)}{\mathrm{a}\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\mathrm{da}\:=_{\mathrm{a}=\mathrm{cos}\theta} \:\:\:\:\int\:\:\frac{\mathrm{arctan}\left(\mathrm{tan}\left(\frac{\theta}{\mathrm{2}}\right)\right)}{\mathrm{cos}\theta\:.\mathrm{sin}\theta}\:\mathrm{sin}\theta\:\mathrm{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\:\:\:\frac{\theta}{\mathrm{cos}\theta}\mathrm{d}\theta…..\mathrm{be}\:\mathrm{continued}…. \\ $$

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