Prove π/2∫0dx1+sin2(tanx)=π2√2(e2+3−2√2e2−3+2√2)\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)

First note that

$$\begin{align} \int_{0}^{\pi /2} \frac{1}{1+ \sin^{2} ( \tan x)} \ dx &= \int_{0}^{\infty} \frac{1}{(1+\sin^{2} t)(1+t^{2})} \ dt \\ &= 2 \int_{0}^{\infty} \frac{1}{(3 - \cos 2t)(1+t^{2})} \ dt \end{align}$$

Then using the identity $$\sum_{k=0}^{\infty} a^{k} \cos(kx) = \frac{1- a \cos x}{1-2a \cos x + a^{2}} , \ \ |a|<1$$

we have

$$1 + 2 \sum_{k=1}^{\infty} a^{k} \cos(kx) = 1 + 2 \left(\frac{1-a \cos x}{1-2a \cos x +a^{2}} -1\right) = \frac{1-a^{2}}{1-2 a \cos x + a^{2}}$$

Therefore,

$$\begin{align} \int_{0}^{\infty} \frac{1}{(1-2a \cos 2x +a^{2})(1+x^{2})} \ dx &= \frac{1}{1-a^{2}} \int_{0}^{\infty} \Big(1+2 \sum_{k=1}^{\infty} a^{k} \cos(2kx) \Big) \ \frac{1}{1+x^{2}} \ dx \\ &= \frac{\pi}{2} \frac{1}{1-a^{2}} + \frac{2}{1-a^{2}} \sum_{k=1}^{\infty} a^{k} \int_{0}^{\infty} \frac{\cos(2kx)}{1+x^{2}} \ dx \\ &= \frac{\pi}{2} \frac{1}{1-a^{2}} + \frac{\pi}{1-a^{2}} \sum_{k=1}^{\infty} \Big(\frac{a}{e^{2}} \Big)^{k} \\ &= \frac{\pi}{2} \frac{1}{1-a^{2}} + \frac{\pi}{1-a^{2}} \frac{a/e^{2}}{1-a^/e^{2}} \\ &= \frac{\pi}{2} \frac{1}{1-a^{2}} \Big(1+ \frac{2a}{e^{2}-a} \Big) = \frac{\pi}{2} \frac{1}{1-a^{2}} \frac{e^{2}+a}{e^{2}-a} \end{align}$$

Now rewrite the integral as

$$\frac{1}{1+a^{2}} \int_{0}^{\infty} \frac{1}{(1 - \frac{2a}{1+a^{2}} \cos 2x)(1+x^{2})} \ dx$$

and let $a= 3 - 2 \sqrt{2}$.

Then

$$\begin{align} \frac{1}{1+a^{2}} \int_{0}^{\infty} \frac{1}{(1- \frac{1}{3} \cos 2x)(1+x^{2})} \ dx &= \frac{3}{1+a^{2}} \int_{0}^{\infty} \frac{1}{(3 - \cos 2x)(1+x^{2})} \ dx \\ &= \frac{\pi}{2} \frac{1}{1-a^{2}} \frac{e^{2} + 3 - 2 \sqrt{2}}{e^{2} - 3 + 2 \sqrt{2}} \end{align}$$

which implies

$$\begin{align} \int_{0}^{\pi /2} \frac{1}{1+\sin^{2} (\tan x)} \ dx &= \frac{\pi}{3} \frac{1+a^{2}}{1-a^{2}}\frac{e^{2} + 3 - 2 \sqrt{2}}{e^{2} - 3 + 2 \sqrt{2}} \\ &= \frac{\pi}{3} \frac{3}{2 \sqrt{2}}\frac{e^{2} + 3 - 2 \sqrt{2}}{e^{2} - 3 + 2 \sqrt{2}} \\ &= \frac{\pi}{2 \sqrt{2}}\frac{e^{2} + 3 - 2 \sqrt{2}}{e^{2} - 3 + 2 \sqrt{2}} \end{align}$$