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Question Number 138771 by qaz last updated on 18/Apr/21

∫_0 ^(π/2) ln (1−tan^2 x+tan^4 x)dx=?

$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\:\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{tan}\:^{\mathrm{4}} {x}\right){dx}=? \\ $$

Answered by Kamel last updated on 19/Apr/21

Answered by MJS_new last updated on 18/Apr/21

∫ln (tan^4 x −tan^2 x +1) dx= [t=tan x → dx=(dt/(t^2 +1))] =∫((ln (t^4 −t^2 +1))/(t^2 +1))dt= [α=(1/2)+((√3)/2)i, β=−(1/2)+((√3)/2)i, γ=α^(−) , δ=β^(−) ] =∫((ln (t−α)+ln (t−β) +ln (t−γ) +ln (t−δ))/(t^2 +1))dt now use ∫((ln (t−y))/(t^2 +1))dt=−(i/2)∫((ln (t−y))/(t−i))dt+(i/2)∫((ln (t−y))/(t+i)) and ∫((ln (t−y))/(t±z))dt=ln ∣x±z∣ ln ∣y±z∣ −Li_2 ((x±z)/(y±z)) ...

$$\int\mathrm{ln}\:\left(\mathrm{tan}^{\mathrm{4}} \:{x}\:−\mathrm{tan}^{\mathrm{2}} \:{x}\:+\mathrm{1}\right)\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}={tan}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\int\frac{\mathrm{ln}\:\left({t}^{\mathrm{4}} −{t}^{\mathrm{2}} +\mathrm{1}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\left[\alpha=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i},\:\beta=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i},\:\gamma=\overline {\alpha},\:\delta=\overline {\beta}\right] \\ $$$$=\int\frac{\mathrm{ln}\:\left({t}−\alpha\right)+\mathrm{ln}\:\left({t}−\beta\right)\:+\mathrm{ln}\:\left({t}−\gamma\right)\:+\mathrm{ln}\:\left({t}−\delta\right)}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\mathrm{now}\:\mathrm{use} \\ $$$$\int\frac{\mathrm{ln}\:\left({t}−{y}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}=−\frac{\mathrm{i}}{\mathrm{2}}\int\frac{\mathrm{ln}\:\left({t}−{y}\right)}{{t}−\mathrm{i}}{dt}+\frac{\mathrm{i}}{\mathrm{2}}\int\frac{\mathrm{ln}\:\left({t}−{y}\right)}{{t}+\mathrm{i}} \\ $$$$\mathrm{and} \\ $$$$\int\frac{\mathrm{ln}\:\left({t}−{y}\right)}{{t}\pm{z}}{dt}=\mathrm{ln}\:\mid{x}\pm{z}\mid\:\mathrm{ln}\:\mid{y}\pm{z}\mid\:−\mathrm{Li}_{\mathrm{2}} \:\frac{{x}\pm{z}}{{y}\pm{z}} \\ $$$$... \\ $$

Commented by qaz last updated on 18/Apr/21

I=∫_0 ^(π/2) ln(1−tan^2 x+tan^4 x)dx =∫_0 ^∞ ((ln(1+x^2 −(√3)x)+ln(1+x^2 +(√3)x))/(1+x^2 ))dx I(a)=∫_0 ^∞ ((ln(1+x^2 −ax)+ln(1+x^2 +ax))/(1+x^2 ))dx I(0)=2∫_0 ^∞ ((ln(1+x^2 ))/(1+x^2 ))dx=2πln2 I(a)′=∫_0 ^∞ (((−x)/(1+x^2 −ax))+(x/(1+x^2 +ax)))(dx/(1+x^2 )) =(2/a)∫_0 ^∞ ((1/(1+x^2 ))−((1+x^2 )/((1+x^2 )^2 −a^2 x^2 )))dx =(2/a)((π/2)−∫_0 ^∞ ((1+x^2 )/(x^4 +(2−a^2 )x^2 +1))dx) =(2/a)((π/2)−∫_0 ^∞ ((1+(1/x^2 ))/(x^2 +(1/x^2 )+(2−a^2 )))dx) =(2/a)((π/2)−∫_0 ^∞ ((d(x−(1/x)))/((x−(1/x))^2 +(4−a^2 )))) =(2/a)((π/2)−(π/( (√(4−a^2 ))))) I=I((√3))=I(0)+∫_0 ^(√3) (2/a)((π/2)−(π/( (√(4−a^2 )))))da=2πln2−πln(4/3)=πln3

$${I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{tan}\:^{\mathrm{4}} {x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}\right)+{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} −{ax}\right)+{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} +{ax}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${I}\left(\mathrm{0}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\mathrm{2}\pi{ln}\mathrm{2} \\ $$$${I}\left({a}\right)'=\int_{\mathrm{0}} ^{\infty} \left(\frac{−{x}}{\mathrm{1}+{x}^{\mathrm{2}} −{ax}}+\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} +{ax}}\right)\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{{a}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }−\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} −{a}^{\mathrm{2}} {x}^{\mathrm{2}} }\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{{a}}\left(\frac{\pi}{\mathrm{2}}−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\left(\mathrm{2}−{a}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\mathrm{1}}{dx}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{{a}}\left(\frac{\pi}{\mathrm{2}}−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\left(\mathrm{2}−{a}^{\mathrm{2}} \right)}{dx}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{{a}}\left(\frac{\pi}{\mathrm{2}}−\int_{\mathrm{0}} ^{\infty} \frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\left(\mathrm{4}−{a}^{\mathrm{2}} \right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{{a}}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\right) \\ $$$${I}={I}\left(\sqrt{\mathrm{3}}\right)={I}\left(\mathrm{0}\right)+\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \frac{\mathrm{2}}{{a}}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\right){da}=\mathrm{2}\pi{ln}\mathrm{2}−\pi{ln}\frac{\mathrm{4}}{\mathrm{3}}=\pi{ln}\mathrm{3} \\ $$

Commented by MJS_new last updated on 18/Apr/21

great! I didn′t think of this

$$\mathrm{great}!\:\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{think}\:\mathrm{of}\:\mathrm{this} \\ $$

Commented by qaz last updated on 18/Apr/21

it′s true

$${it}'{s}\:{true} \\ $$

Commented by MJS_new last updated on 18/Apr/21

yes

$$\mathrm{yes} \\ $$

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