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$u=\dfrac{π}{2}-x$$Ι=\int_{0}^{\dfrac{\pi}{2}}\dfrac{e^{\sqrt{2}lntanu}}{1+e^{\sqrt{2}lntanu}}du$=>$2I=\dfrac{π}{2}$=>$Ι=\dfrac{π}{4}$.
$u=\dfrac{π}{2}-x$
ΑπάντησηΔιαγραφή$Ι=\int_{0}^{\dfrac{\pi}{2}}\dfrac{e^{\sqrt{2}lntanu}}{1+e^{\sqrt{2}lntanu}}du$=>$2I=\dfrac{π}{2}$=>
$Ι=\dfrac{π}{4}$.