计算下面两个积分的比值：
[Largedisplaystyle int_{0}^{1}frac{1}{sqrt{1+t^{4}}}mathrm{d}t~,~int_{0}^{1}frac{1}{sqrt{1-t^{4}}}mathrm{d}t
]
(Largemathbf{Solution:})
由
[egin{align*}
int_{0}^{1}frac{1}{sqrt{1+t^{4}}}mathrm{d}t &= int_{0}^{1}frac{mathrm{d}t}{sqrt{left ( 1+t^{2}
ight )^{2}-2t^{2}}}\
&=frac{1}{2}int_{0}^{1}frac{1}{sqrt{1-dfrac{1}{2}left ( dfrac{2t}{1+t^{2}}
ight )^{2}}}frac{2}{1+t^{2}}mathrm{d}t\
left ( t= anfrac{ heta }{2}
ight ) &=frac{1}{2}int_{0}^{frac{pi }{2}}frac{mathrm{d} heta }{sqrt{1-dfrac{1}{2}sin^{2} heta }}=frac{1}{2}mathbf{K}left ( frac{1}{sqrt{2}}
ight )
end{align*}]
和
[egin{align*}
int_{0}^{1}frac{1}{sqrt{1-t^{4}}}mathrm{d}t&=int_{0}^{frac{pi }{2}}frac{mathrm{d} heta }{sqrt{1+cos^{2} heta }} left ( t=cos heta
ight )\
&=int_{0}^{frac{pi }{2}}frac{mathrm{d} heta }{sqrt{2-sin^{2} heta }}\
&=frac{1}{sqrt{2}}mathbf{K}left ( frac{1}{sqrt{2}}
ight )
end{align*}]
我们可以得到
[Largeoxed{displaystyle {frac{displaystyle int_{0}^{1}frac{1}{sqrt{1-x^{4}}}mathrm{d}x}{displaystyle int_{0}^{1}frac{1}{sqrt{1+x^{4}}}mathrm{d}x}}=frac{dfrac{1}{sqrt{2}}mathbf{K}left ( dfrac{1}{sqrt{2}}
ight )}{dfrac{1}{2}mathbf{K}left ( dfrac{1}{sqrt{2}}
ight )}=color{blue}{sqrt {2}}}
]