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Question Number 137155 by mnjuly1970 last updated on 30/Mar/21

......nice calculus ...... evaluate :: 𝛗=∫_0 ^( 2Ο€) (1/(1+cos^4 (x)))dx=???

$$\:\:\:\:\:\:\:\:......{nice}\:\:\:\:{calculus}\:...... \\ $$$$\:\:\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \frac{\mathrm{1}}{\mathrm{1}+{cos}^{\mathrm{4}} \left({x}\right)}{dx}=??? \\ $$

Answered by Ar Brandon last updated on 30/Mar/21

Ο†=∫_0 ^(2Ο€) (dx/(1+cos^4 x))=4∫_0 ^(Ο€/2) ((sec^4 x)/(sec^4 x+1))dx , t=tanx =4∫_0 ^∞ ((t^2 +1)/((t^2 +1)^2 +1))dt=4∫_0 ^∞ ((t^2 +1)/(t^4 +2t^2 +2))dt =2∫_0 ^∞ (((t^2 +(√2))+(t^2 βˆ’(√2)))/(t^4 +2t^2 +2))dt+(2/( (√2)))∫_0 ^∞ (((t^2 +(√2))βˆ’(t^2 βˆ’(√2)))/(t^4 +2t^2 +2))dt =2∫_0 ^∞ {(((1+((√2)/t^2 ))+(1βˆ’((√2)/t^2 )))/(t^2 +2+(2/t^2 )))}dt+(2/( (√2)))∫_0 ^∞ {(((1+((√2)/t^2 ))βˆ’(1βˆ’((√2)/t^2 )))/(t^2 +2+(2/t^2 )))}dt =2∫_0 ^∞ {((1+((√2)/t^2 ))/((tβˆ’((√2)/t))^2 +2(√2)+2))+((1βˆ’((√2)/t^2 ))/((t+((√2)/t))^2 +2βˆ’2(√2)))}dt +(2/( (√2)))∫_0 ^∞ {((1+((√2)/t^2 ))/((tβˆ’((√2)/t))^2 +2(√2)+2))βˆ’((1βˆ’((√2)/t^2 ))/((t+((√2)/t))^2 +2βˆ’2(√2)))}dt =2∫_(βˆ’βˆž) ^(+∞) (du/(u^2 +2(√2)+2))+(2/( (√2)))∫_(βˆ’βˆž) ^(+∞) (dv/(v^2 +2(√2)+2)) =[(2/( (√(2(√2)+2))))tan^(βˆ’1) ((u/( (√(2(√2)+2)))))+((√2)/( (√(2(√2)+2))))tan^(βˆ’1) ((v/( (√(2(√2)+2)))))]_(βˆ’βˆž) ^(+∞) =((2Ο€)/( (√(2(√2)+2))))+(((√2)Ο€)/( (√(2(√2)+2))))

$$\phi=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{cos}^{\mathrm{4}} \mathrm{x}}=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sec}^{\mathrm{4}} \mathrm{x}}{\mathrm{sec}^{\mathrm{4}} \mathrm{x}+\mathrm{1}}\mathrm{dx}\:,\:\mathrm{t}=\mathrm{tanx} \\ $$$$\:\:\:=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\mathrm{dt}=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{t}^{\mathrm{4}} +\mathrm{2t}^{\mathrm{2}} +\mathrm{2}}\mathrm{dt} \\ $$$$\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}\right)+\left(\mathrm{t}^{\mathrm{2}} βˆ’\sqrt{\mathrm{2}}\right)}{\mathrm{t}^{\mathrm{4}} +\mathrm{2t}^{\mathrm{2}} +\mathrm{2}}\mathrm{dt}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}\right)βˆ’\left(\mathrm{t}^{\mathrm{2}} βˆ’\sqrt{\mathrm{2}}\right)}{\mathrm{t}^{\mathrm{4}} +\mathrm{2t}^{\mathrm{2}} +\mathrm{2}}\mathrm{dt} \\ $$$$\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \left\{\frac{\left(\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{t}^{\mathrm{2}} }\right)+\left(\mathrm{1}βˆ’\frac{\sqrt{\mathrm{2}}}{\mathrm{t}^{\mathrm{2}} }\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{2}}{\mathrm{t}^{\mathrm{2}} }}\right\}\mathrm{dt}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} \left\{\frac{\left(\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{t}^{\mathrm{2}} }\right)βˆ’\left(\mathrm{1}βˆ’\frac{\sqrt{\mathrm{2}}}{\mathrm{t}^{\mathrm{2}} }\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{2}}{\mathrm{t}^{\mathrm{2}} }}\right\}\mathrm{dt} \\ $$$$\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \left\{\frac{\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{t}βˆ’\frac{\sqrt{\mathrm{2}}}{\mathrm{t}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}+\frac{\mathrm{1}βˆ’\frac{\sqrt{\mathrm{2}}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{t}+\frac{\sqrt{\mathrm{2}}}{\mathrm{t}}\right)^{\mathrm{2}} +\mathrm{2}βˆ’\mathrm{2}\sqrt{\mathrm{2}}}\right\}\mathrm{dt} \\ $$$$\:\:\:\:\:+\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} \left\{\frac{\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{t}βˆ’\frac{\sqrt{\mathrm{2}}}{\mathrm{t}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}βˆ’\frac{\mathrm{1}βˆ’\frac{\sqrt{\mathrm{2}}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{t}+\frac{\sqrt{\mathrm{2}}}{\mathrm{t}}\right)^{\mathrm{2}} +\mathrm{2}βˆ’\mathrm{2}\sqrt{\mathrm{2}}}\right\}\mathrm{dt} \\ $$$$\:\:\:\:=\mathrm{2}\int_{βˆ’\infty} ^{+\infty} \frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}\int_{βˆ’\infty} ^{+\infty} \frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}} \\ $$$$\:\:\:\:=\left[\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}}\mathrm{tan}^{βˆ’\mathrm{1}} \left(\frac{\mathrm{u}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}}\right)+\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}}\mathrm{tan}^{βˆ’\mathrm{1}} \left(\frac{\mathrm{v}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}}\right)\right]_{βˆ’\infty} ^{+\infty} \\ $$$$\:\:\:\:=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}}+\frac{\sqrt{\mathrm{2}}\pi}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}} \\ $$

Commented by Ar Brandon last updated on 30/Mar/21

Not 100% sure... Please check

$$\mathrm{Not}\:\mathrm{100\%}\:\mathrm{sure}...\:\mathrm{Please}\:\mathrm{check} \\ $$

Commented by Dwaipayan Shikari last updated on 30/Mar/21

((2+(√2))/( (√(2(√2)+2))))Ο€=(√(1+(√2))) Ο€

$$\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}}\pi=\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}\:\pi\:\: \\ $$

Commented by mnjuly1970 last updated on 30/Mar/21

thanks alot mr brandon..

$$\:{thanks}\:{alot}\:{mr}\:{brandon}.. \\ $$

Commented by mnjuly1970 last updated on 30/Mar/21

grateful mr payan...

$${grateful}\:{mr}\:{payan}... \\ $$

Commented by Ar Brandon last updated on 30/Mar/21

Cool

Answered by mathmax by abdo last updated on 31/Mar/21

Ξ¦=∫_0 ^(2Ο€) (dx/(1+cos^4 x)) β‡’Ξ¦ =∫_0 ^(2Ο€) (dx/(1+(((1+cos(2x))/2))^2 )) =∫_0 ^(2Ο€) ((4dx)/(4+(1+2cos(2x)+cos^2 (2x)))) =4∫_0 ^(2Ο€) (dx/(5 +2cos(2x)+((1+cos(4x))/2))) =8 ∫_0 ^(2Ο€) (dx/(10+4cos(2x)+1+cos(4x))) =8 ∫_0 ^(2Ο€) (dx/(11+4cos(2x)+cos(4x))) =_(2x=t) 8∫_0 ^(4Ο€) (dt/(2(11+4cost +cos(2t)))) =4 ∫_0 ^(4Ο€) (dt/(11+4cost +2cos^2 tβˆ’1)) =4 ∫_0 ^(4Ο€) (dt/(2cos^2 t+4cost +10)) =2∫_0 ^(4Ο€) (dt/(cos^2 t+2cost +5)) z^2 +2z+5=0β†’Ξ”^β€² =1βˆ’5=βˆ’4 β‡’z_1 =βˆ’1+2i and z_2 =βˆ’1βˆ’2i β‡’ cos^2 t+2cost+5 =(costβˆ’z_1 )(costβˆ’z_2 ) β‡’ (1/(cos^2 t+2cost+5)) =(1/((costβˆ’z_1 )(costβˆ’z_2 ))) =((1/(costβˆ’z_1 ))βˆ’(1/(costβˆ’z_2 )))Γ—(1/(4i)) β‡’ Ξ¦ =(1/(2i))∫_0 ^(4Ο€) (dt/(costβˆ’z_1 ))βˆ’(1/(2i))∫_0 ^(4Ο€) (dt/(costβˆ’z_2 )) =βˆ’i∫_0 ^(2Ο€) (dt/(costβˆ’z_1 )) +i∫_0 ^(2Ο€) (dt/(costβˆ’z_2 )) =βˆ’iH +iK H =_(z=e^(it) ) ∫_(∣z∣=1) (dz/(iz(((z+z^(βˆ’1) )/2)+1βˆ’2i))) =∫_(∣z∣=1) ((2dz)/(iz(z+z^(βˆ’1) +2βˆ’4i))) =∫ ((βˆ’2i)/(z^2 +1+(2βˆ’4i)z)) Ο•(z)=((βˆ’2i)/(z^2 +(2βˆ’4i)z +1)) poles? Ξ”^β€² =(1βˆ’2i)^2 βˆ’1 =1βˆ’4iβˆ’4βˆ’1 =βˆ’4(1+i) =βˆ’4(√2)e^((iΟ€)/4) β‡’ (βˆšΞ”^, )=2i(^4 (√2)) e^((iΟ€)/8) β‡’z_1 =βˆ’1+2i +2i(^4 (√2))e^((iΟ€)/8) =βˆ’1+2i +2i(^4 (√2))(((√(2+(√2)))/2)+i((√(2βˆ’(√2)))/2)) =βˆ’1+2i+2i(^4 (√2))((√(2+(√2)))/2) βˆ’(^4 (√2))(√(2βˆ’(√2))) =βˆ’1+2i+i(^4 (√2))(√(2+(√2)))βˆ’(^4 (√2))(√(2βˆ’(√2))) β‡’βˆ£z_1 ∣=(√((1+(^4 (√2))(√(2βˆ’(√2))))^2 +(√2)(2+(√2)))) ....be continued....

$$\Phi=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{cos}^{\mathrm{4}} \mathrm{x}}\:\Rightarrow\Phi\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{dx}}{\mathrm{1}+\left(\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{4dx}}{\mathrm{4}+\left(\mathrm{1}+\mathrm{2cos}\left(\mathrm{2x}\right)+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{2x}\right)\right)}\:=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{dx}}{\mathrm{5}\:+\mathrm{2cos}\left(\mathrm{2x}\right)+\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{4x}\right)}{\mathrm{2}}} \\ $$$$=\mathrm{8}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{dx}}{\mathrm{10}+\mathrm{4cos}\left(\mathrm{2x}\right)+\mathrm{1}+\mathrm{cos}\left(\mathrm{4x}\right)}\:=\mathrm{8}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{dx}}{\mathrm{11}+\mathrm{4cos}\left(\mathrm{2x}\right)+\mathrm{cos}\left(\mathrm{4x}\right)} \\ $$$$=_{\mathrm{2x}=\mathrm{t}} \:\:\:\mathrm{8}\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\frac{\mathrm{dt}}{\mathrm{2}\left(\mathrm{11}+\mathrm{4cost}\:+\mathrm{cos}\left(\mathrm{2t}\right)\right)}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\frac{\mathrm{dt}}{\mathrm{11}+\mathrm{4cost}\:+\mathrm{2cos}^{\mathrm{2}} \mathrm{t}βˆ’\mathrm{1}} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\frac{\mathrm{dt}}{\mathrm{2cos}^{\mathrm{2}} \mathrm{t}+\mathrm{4cost}\:+\mathrm{10}}\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\frac{\mathrm{dt}}{\mathrm{cos}^{\mathrm{2}} \mathrm{t}+\mathrm{2cost}\:+\mathrm{5}} \\ $$$$\mathrm{z}^{\mathrm{2}} \:+\mathrm{2z}+\mathrm{5}=\mathrm{0}\rightarrow\Delta^{'} \:=\mathrm{1}βˆ’\mathrm{5}=βˆ’\mathrm{4}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =βˆ’\mathrm{1}+\mathrm{2i}\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =βˆ’\mathrm{1}βˆ’\mathrm{2i}\:\Rightarrow \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{t}+\mathrm{2cost}+\mathrm{5}\:=\left(\mathrm{cost}βˆ’\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{cost}βˆ’\mathrm{z}_{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{t}+\mathrm{2cost}+\mathrm{5}}\:=\frac{\mathrm{1}}{\left(\mathrm{cost}βˆ’\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{cost}βˆ’\mathrm{z}_{\mathrm{2}} \right)} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{cost}βˆ’\mathrm{z}_{\mathrm{1}} }βˆ’\frac{\mathrm{1}}{\mathrm{cost}βˆ’\mathrm{z}_{\mathrm{2}} }\right)Γ—\frac{\mathrm{1}}{\mathrm{4i}}\:\Rightarrow \\ $$$$\Phi\:=\frac{\mathrm{1}}{\mathrm{2i}}\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\frac{\mathrm{dt}}{\mathrm{cost}βˆ’\mathrm{z}_{\mathrm{1}} }βˆ’\frac{\mathrm{1}}{\mathrm{2i}}\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\frac{\mathrm{dt}}{\mathrm{cost}βˆ’\mathrm{z}_{\mathrm{2}} } \\ $$$$=βˆ’\mathrm{i}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{dt}}{\mathrm{cost}βˆ’\mathrm{z}_{\mathrm{1}} }\:+\mathrm{i}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{dt}}{\mathrm{cost}βˆ’\mathrm{z}_{\mathrm{2}} }\:=βˆ’\mathrm{iH}\:+\mathrm{iK} \\ $$$$\mathrm{H}\:=_{\mathrm{z}=\mathrm{e}^{\mathrm{it}} } \:\:\:\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{dz}}{\mathrm{iz}\left(\frac{\mathrm{z}+\mathrm{z}^{βˆ’\mathrm{1}} }{\mathrm{2}}+\mathrm{1}βˆ’\mathrm{2i}\right)} \\ $$$$=\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\frac{\mathrm{2dz}}{\mathrm{iz}\left(\mathrm{z}+\mathrm{z}^{βˆ’\mathrm{1}} \:+\mathrm{2}βˆ’\mathrm{4i}\right)}\:=\int\:\frac{βˆ’\mathrm{2i}}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}+\left(\mathrm{2}βˆ’\mathrm{4i}\right)\mathrm{z}} \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{βˆ’\mathrm{2i}}{\mathrm{z}^{\mathrm{2}} \:+\left(\mathrm{2}βˆ’\mathrm{4i}\right)\mathrm{z}\:+\mathrm{1}}\:\:\mathrm{poles}? \\ $$$$\Delta^{'} \:=\left(\mathrm{1}βˆ’\mathrm{2i}\right)^{\mathrm{2}} βˆ’\mathrm{1}\:=\mathrm{1}βˆ’\mathrm{4i}βˆ’\mathrm{4}βˆ’\mathrm{1}\:=βˆ’\mathrm{4}\left(\mathrm{1}+\mathrm{i}\right)\:=βˆ’\mathrm{4}\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\sqrt{\Delta^{,} }=\mathrm{2i}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{8}}} \:\Rightarrow\mathrm{z}_{\mathrm{1}} =βˆ’\mathrm{1}+\mathrm{2i}\:+\mathrm{2i}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{8}}} \\ $$$$=βˆ’\mathrm{1}+\mathrm{2i}\:+\mathrm{2i}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{2}βˆ’\sqrt{\mathrm{2}}}}{\mathrm{2}}\right) \\ $$$$=βˆ’\mathrm{1}+\mathrm{2i}+\mathrm{2i}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\:βˆ’\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\sqrt{\mathrm{2}βˆ’\sqrt{\mathrm{2}}} \\ $$$$=βˆ’\mathrm{1}+\mathrm{2i}+\mathrm{i}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}βˆ’\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\sqrt{\mathrm{2}βˆ’\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\mid\mathrm{z}_{\mathrm{1}} \mid=\sqrt{\left(\mathrm{1}+\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\sqrt{\mathrm{2}βˆ’\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}\:....\mathrm{be}\:\mathrm{continued}.... \\ $$

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