<s>[INST] Prove $\int\limits_{0}^{\pi/2}\frac{dx}{1+\sin^2{(\tan{x})}}=\frac{\pi}{2\sqrt{2}}\bigl(\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\bigr)$ 
Prove the following integral
  $$I=\int\limits_{0}^{\frac{\pi}{2}}\dfrac{dx}{1+\sin^2{(\tan{x})}}=\dfrac{\pi}{2\sqrt{2}}\left(\dfrac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\right)$$

This integral result was calculated using Mathematica and 
I like this integral. But I can't solve it. 
My idea:
Let $$\tan{x}=t\Longrightarrow dx=\dfrac{1}{1+t^2}dt$$
so
$$I=\int\limits_{0}^{\infty}\dfrac{dt}{1+\sin^2{t}}\cdot \dfrac{1}{1+t^2}$$
then I can't proceed. Can you help me? Thank you.
 [/INST] I hope nobody cares that i exhume this question, but i found it interesting that it is possible to obtain this integral by a relativly straightforward contour integration method.
Observe that,following the question opener and using parity, that we can rewrite the integral as
$$
\frac{1}{2}\int^{\infty}_{-\infty}\frac{1}{1+t^2}\frac{1}{1+\sin^2(t)}
$$
It is now easy to show that the poles are
$$
t_{\pm}=\pm i\\
t_{n\pm}=\pi n\pm i \text{arcsinh(1)}
$$
so we have two isolated poles and the rest lies on two straight lines paralell to the real axis.
Because the integrand interpreted as a complex function converges as $|z|\rightarrow\infty$ we can choose a semicircle closed in the upper half  plane as an integration contour. We find
$$
I=\pi i\sum_{n=-\infty}^{\infty}\text{res}(t_{n+})+\pi i \text{res}(t_{+})
$$
Where the residues are given by
$$
\text{res}(t_{+})=\frac{i}{2}\frac{1}{2 \sinh^2(1)-1}\\
\text{res}(t_{n+})=\frac{-i}{2\sqrt{2}}\frac{1}{1+(n \pi+i \text{arcsinh(1)} )^2}
$$
Therefore the integral reduces to the following sum
$$
I=\frac{\pi}{2\sqrt{2}} \sum_{n=-\infty}^{\infty} \frac{1}{1+(n \pi+i \text{arcsinh(1)})^2} -\frac{\pi}{2}\frac{1}{2 \sinh^2(1)-1}
$$
Using a partial fraction decomposition together with the Mittag-Leffler expansion of $\coth(x)$, this can be rewritten as
$$
I=\frac{\pi}{4\sqrt{2}} \sum_{n=-\infty}^{\infty} \frac{-i}{-i+n \pi+ \text{arcsinh(1)}}+ \frac{i}{i+n \pi+ i\text{arcsinh(1)}}-\frac{\pi}{2}\frac{1}{2 \sinh^2(1)-1}=\\
\frac{\sqrt{2} \pi}{8} \left(  \coth \left(1-\text{arcsinh(1)}\right)+  \coth \left(1+\text{arcsinh(1)}\right)\right)-\frac{\pi}{2}\frac{1}{2 \sinh^2(1)-1}\\
$$
Or 
$$
I\approx 1.16353
$$
Which matches the claimed result.
One can also compute this explicitly noting that $\text{arcsinh(1)}=\log(1+\sqrt{2})$ (*). But this is rather tedious so i just leave this step to the reader and conclude that
$$
I=\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}
$$
Appendix
Just to give some details of the last part of the calculations:
Using (*) the part stemming from the sum is 
$$
\frac{\pi}{4\sqrt{2}}\left(\frac{ \frac{1+\sqrt{2}}{e}+\frac{e}{1+\sqrt{2}}}{ \frac{e}{1+\sqrt{2}}-\frac{1+\sqrt{2}}{e}}+\frac{e \left(1+\sqrt{2}\right)+\frac{1}{1+\sqrt{2} e} }{\left(1+\sqrt{2}\right) e-\frac{1}{\left(1+\sqrt{2}\right) e}}\right)=\\
\frac{\left(e^4-1\right) \pi }{2 \sqrt{2} \left(1-6 e^2+e^4\right)}
$$
The part of the single pole gives
$$
\frac{\pi }{2 \left(\left(\frac{e}{2}-\frac{1}{2 e}\right)^2-1\right)}=\frac{2 e^2 \pi }{1-6 e^2+e^4}
$$
Adding both terms and factorizing then yields the desired result
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