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Question Number 154080 by iloveisrael last updated on 14/Sep/21
Ω =∫_0 ^( (π/2)) ln^2 (((1+sin t)/(1−sin t)))dt
$$\:\:\:\:\Omega\:=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:^{\mathrm{2}} \left(\frac{\mathrm{1}+\mathrm{sin}\:{t}}{\mathrm{1}−\mathrm{sin}\:{t}}\right){dt} \\ $$
Answered by mindispower last updated on 14/Sep/21
=∫_0 ^(π/2) 4ln^2 (((1+tg((x/2)))/(1−tg((x/2)))))dx y=tg((x/2))⇒dx=((2dy)/(1+y^2 )) =8∫_0 ^1 ((ln^2 (((1−y)/(1+y))))/(1+y^2 ))dy y=((1−x)/(1+x))⇒dy=((−2dx)/((1+x)^2 )) Ω=8∫_0 ^1 ((ln^2 (x))/(1+(((1−x)/(1+x)))^2 )).((2dx)/((1+x)^2 ))=8∫_0 ^1 ((ln^2 (x))/(1+x^2 ))dx =8∫_0 ^1 Σ_(n≥0) (−x^2 )^n ln^2 (x)dx =8Σ_(n≥0) (−1)^n ∫_0 ^1 x^(2n) ln^2 (x)dx ∫_0 ^1 x^(2n) ln^2 (x)dx=∫_0 ^∞ t^2 e^(−(2n+1)t) dt (1/((2n+1)^3 ))∫_0 ^∞ t^2 e^(−t) =(2/((1+2n)^3 )) Ω=16Σ_(n≥0) (((−1)^n )/((1+2n)^3 )) =16Σ_(n≥0) ((1/((1+4n)^3 ))−(1/((4n+3)^3 ))) =((16)/(64))Σ_(n≥0) ((1/((n+(1/4))^3 ))−(1/((n+(3/4))^3 ))) =((16)/(64))(Ψ^3 ((1/4))−Ψ^3 ((3/4)))
$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{4}{ln}^{\mathrm{2}} \left(\frac{\mathrm{1}+{tg}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}−{tg}\left(\frac{{x}}{\mathrm{2}}\right)}\right){dx} \\ $$$${y}={tg}\left(\frac{{x}}{\mathrm{2}}\right)\Rightarrow{dx}=\frac{\mathrm{2}{dy}}{\mathrm{1}+{y}^{\mathrm{2}} } \\ $$$$=\mathrm{8}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\frac{\mathrm{1}−{y}}{\mathrm{1}+{y}}\right)}{\mathrm{1}+{y}^{\mathrm{2}} }{dy} \\ $$$${y}=\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\Rightarrow{dy}=\frac{−\mathrm{2}{dx}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} } \\ $$$$\Omega=\mathrm{8}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{2}} }.\frac{\mathrm{2}{dx}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }=\mathrm{8}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{8}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}\geqslant\mathrm{0}} {\sum}\left(−{x}^{\mathrm{2}} \right)^{{n}} {ln}^{\mathrm{2}} \left({x}\right){dx} \\ $$$$=\mathrm{8}\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}} {ln}^{\mathrm{2}} \left({x}\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}} {ln}^{\mathrm{2}} \left({x}\right){dx}=\int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{2}} {e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){t}} {dt} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{2}} {e}^{−{t}} =\frac{\mathrm{2}}{\left(\mathrm{1}+\mathrm{2}{n}\right)^{\mathrm{3}} } \\ $$$$\Omega=\mathrm{16}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{1}+\mathrm{2}{n}\right)^{\mathrm{3}} } \\ $$$$=\mathrm{16}\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{4}{n}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{4}{n}+\mathrm{3}\right)^{\mathrm{3}} }\right) \\ $$$$=\frac{\mathrm{16}}{\mathrm{64}}\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} }\right) \\ $$$$=\frac{\mathrm{16}}{\mathrm{64}}\left(\Psi^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\Psi^{\mathrm{3}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right) \\ $$$$ \\ $$$$ \\ $$

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