Find the range of: $y=\sqrt{\sin(\log_e\frac{x^2+e}{x^2+1})}+\sqrt{\cos(\log_e\frac{x^2+e}{x^2+1})}$ Find the range of:
$$y=\sqrt{\sin(\log_e\frac{x^2+e}{x^2+1})}+\sqrt{\cos(\log_e\frac{x^2+e}{x^2+1})}$$
What I tried:
Let:$$\log_e\frac{x^2+e}{x^2+1}=X,$$
then $$y=\sqrt {\sin X}+\sqrt{\cos X}$$
$$y_{max}at X=\pi/4$$
The rest is too complicated. I am stuck up.
Can someone please give a analytical solution.
Even Wolfram alpha doesn't help.
 A: A related problem. First, we study the expression 
$$ \frac{x^2+e}{x^2+1}=1+\frac{e-1}{x^2+1} \longrightarrow_{|x|\to\infty} 1,  $$
which implies $ y(x)\longrightarrow_{|x|\to \infty} 0 $. To find the maximum of the function $y(x)$, Let's study the function
$$ h(t)=\sqrt{\sin(t)}+\sqrt{\cos(t)}.  $$
The maximum of the above function is attained at $t=\frac{\pi}{4}$ which can proved using derivative test. So, this implies our function attains its max when
$$ \ln\left( \frac{x^2+e}{x^2+1} \right)=\frac{\pi}{4} \implies x=0.6632987771, -0.6632987771. $$
Plugging back in the function y(x) gives the max which is $y=1.681792830$. So the range is
$$ 1 < y \leq 1.681792830.$$
Note: You can solve $ \ln\left( \frac{x^2+e}{x^2+1} \right)=\frac{\pi}{4} $ easily as,
$$ \ln\left( \frac{x^2+e}{x^2+1} \right)=\frac{\pi}{4}\implies \frac{x^2+e}{x^2+1} =e^{\frac{\pi}{4}}\implies x^2+e= e^{\frac{\pi}{4}}x^2+ e^{\frac{\pi}{4}}=\dots.$$
I think you can finish it.

A: let $f(x)=\frac{x^2+e}{x^2+1} \to 1<f(x)\le e \to 0<X\le 1$
$y^2=sinX+cosX+\sqrt{2sin2X}=\sqrt{2}\left(sin(X+\dfrac{\pi}{4})+\sqrt{sin2X}\right)$
when $0<X<\dfrac{\pi}{4}$,$sin2X$ and $sin(X+\dfrac{\pi}{4})$ are mono increasing,$\to y^2>1 \to y>1$
I left another part for you to discuss and you should be able to find out the range now. 
