Evaluating $\left.\int_0^{\pi/2}\sqrt{1+\frac1{\sqrt{1+\tan^nx}}}\text dx \middle/\int_0^{\pi/2}\sqrt{1-\frac1{\sqrt{1+\tan^nx}}}\text dx\right.$ 
Evaluate the integral ratio$$\dfrac{I_1}{I_2}=\dfrac{\displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{1+\frac{1}{\sqrt{1+\tan^nx}}} \mathrm{d}x}{\displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}} \mathrm{d}x}$$

Using Wolframalpha, I found out that the ratio of the above two integrals is $\sqrt{2}+1$, that is, independent of $n$. I couldn't think of a way to solve it.
I don't think any simple substitutions would work here. Any useful hints would be appreciated.

Note
\begin{align}
\frac{I_1}{I_2}&=1+ \frac{I_1-I_2}{I_2} \\
&=1+\frac1{I_2}\int_0^{\frac\pi2} \left( \sqrt{1+\frac{1}{\sqrt{1+\tan^nx}}}-  \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}} \right)dx \\
&=1+\frac1{I_2}\int_0^{\frac\pi2}
\frac{\sqrt{\sqrt{\sin^nx+\cos^nx}+\sqrt{\cos^nx}}
 - \sqrt{\sqrt{\sin^nx+\cos^nx}-\sqrt{\cos^nx}}}{\sqrt[4]{\sin^nx+\cos^nx}}dx \\
&=1+\frac1{I_2} \int_0^{\frac\pi2}
\frac{\sqrt{2\sqrt{\sin^nx+\cos^nx}-2\sqrt{\sin^nx}}
}{\sqrt[4]{\sin^nx+\cos^nx}}dx \\
&= 1+\frac{\sqrt2}{I_2} \int_0^{\frac\pi2} \sqrt{1-\frac{1}{\sqrt{1+\cot^nx}}}dx\>\>\>\>\>\>\>\>(x\to \frac\pi2-x)\\
&=1+\frac{\sqrt2}{I_2}\int_0^{\frac\pi2} \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}}dx \\ &=1+\sqrt2 \\
\end{align}
