Verified answer

Siga o passo a passo abaixo:

[tex]\large\sf{I=\displaystyle\int^{\sf{\frac{\pi}{2}}}_{0} \sf{\cfrac{1}{1+\sin^4(x)+\cos^4(x)}}\:\:dx}[/tex]

[tex]\large\sf{I=\displaystyle\int^{\sf{\frac{\pi}{2}}}_{0}\sf{\cfrac{1}{1+(\sin^2(x) + \cos^2(x))^2-2\sin^2(x)\cos^2(x)}}}[/tex]

[tex]\large\sf{I=\displaystyle\int^{\sf{\frac{\pi}{2}}}_{0}\sf{\cfrac{1}{2-\frac{1}{2}\sin^2(2x)}}\:\:dx\:\:=\:\:2\displaystyle\int^{\sf{\frac{\pi}{2}}}_{0}\sf{\cfrac{1}{4-\sin^2(2x)}}}[/tex]

[tex]\large\sf{I=\displaystyle\int^{\sf{\frac{\pi}{2}}}_{0}}\cfrac{1}{4-\left( \cfrac{2\tan(x)}{1+\tan^2(x)}\right)^2}\:\:dx\:\:=\:\:2\displaystyle\int^{\sf{\cfrac{\pi}{2}}}_0\sf{\cfrac{(1+\tan^2(x))^2}{4+4\tan^2(x)+4\tan^4(x)}}\:\: dx[/tex]

[tex]\large\sf{I=\cfrac{1}{2}\displaystyle\int^{\sf{+\infty}}_{0}\sf{\frac{1+u^2}{1+u^2+u^4}\:\:du\:\:=\:\:\sf{\frac{1}{2}\displaystyle\int^{\sf{+\infty}}_{0}\sf{\cfrac{1+u^{-2}}{(u-u^{-1})^2+3}}}}\:\: du}[/tex]

[tex]\large\sf{I=\frac{1}{2}\displaystyle\int^{\sf{+\infty}}_{- \infty}\sf{\frac{dv}{v^2+3}}\:\:=\:\:\cfrac{1}{2}\left\{ \cfrac{1}{\sqrt{3}}\tan^{-1}\left( \cfrac{v}{\sqrt{3}}\right)\right\}^{+\infty}_{-\infty}}[/tex]

[tex]\large\sf{I=\cfrac{1}{2\sqrt{3}}\left( \cfrac{\pi}{2}-\left(-\cfrac{\pi}{2} \right)\right)\:\:=\: \:\cfrac{\pi}{2\sqrt{3}}}[/tex]

[tex] \red{ \boxed{ \therefore \: \:\large{ \sf{I = \displaystyle\int ^{ \large{ \sf{ \cfrac{\pi}{2} } } } _{0} \cfrac{ \sf{1}}{ \sf{1 + { \sin}^{4}(x) + { \cos }^{4} (x) }} } \: \: \: dx \: \: = \: \: \cfrac{\pi}{2 \sqrt{3} }} }}[/tex]

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