Ok, there is good way to proof this sequence is bounded.
From##a_{n}=\sqrt{2a_{n-1}}##
##a_{n}<2## as ## a_{n-1}<2##
However we know##a_{1}=\sqrt{2}<2##
So we kown##a_{2}<2... a_{n}<2##
First, we can know
##a_{n}=\sqrt{2a_{n-1}}##
When##n\rightarrow \infty##
##a_{n}=\sqrt{2a_{n-1}}##
And we can get the answer is 2.
Is this solution right? And is any other way to solve the question?
Maybe I have found a better way to solve the question. ##a_n=\int_{0}^{2-\sqrt{3}}\frac{1-x^{4n}}{1+x^{2}}dx=\int_{0}^{2-\sqrt{3}}\frac{1}{1+x^{2}}dx-\int_{0}^{2-\sqrt{3}}\frac{x^{4n}}{1+x^{2}}dx=arctan(2-\sqrt{3})...
In my opinion , if it can be shown that this is a monotonically bounded sequence, one can confirm that there is a limit.
First,we know $$ \frac{1-x^{4n}}{1+x^{2}}dx=(1-x^{2}) (1+x^{2}) ^{n-1}=(1-x^{4}) ^{n-1}(1+x^{2}).$$
According to the integral median theorem,we can get $$a_n=(2- \sqrt{3} )...
\frac{d^{2}}{dx^{2}}\int_{0}^{x}(\int_{1}^{sint}\sqrt{1+u^{4}}du)dt=\frac{d}{dx}\int_{0}^{sinx}(\sqrt{1+u^{4}})du
then we let m=sinx,so x=arcsinx,then we get \frac{d}{dx}\int_{0}^{sinx}(\sqrt{1+u^{4}})du=\frac{dm}{dx}\frac{d}{dm}\int_{0}^{m}(\sqrt{1+u^{4}})du=\sqrt{1+m^{4}}\frac{dm}{dx},then we...
When I encountereD this kind of question before.For example
\int x\sqrt{2+x^{2}}dx
We make the Substitution t=x^{2}+2,because its differential is dt=2xdx,so we get \int x\sqrt{2+x^{2}}=1/2\int\sqrt{t}dt,then we can get the answer easily
But the question,it seems that I can't use the way to...