How to solve $\int _{\frac{\pi }{8}}^{\frac{\pi }{3}}\frac{dx}{1+\left( \tan x\right) ^{\sqrt{2}}}$ Calculate $$\int _{\pi/8}^{3\pi/8}\dfrac{dx}{1+\left( \tan x\right) ^{\sqrt{2}}}\,.$$

I thought.
If I put $t=\tan x$, it becomes $dx=\cos ^{2}xdt$ and $x$ remains.
So I can do integration by substitution.
And I thought that rationalize the denominator but I can't do.
This problem looks to be a trap.
If you look at the integrand in the range of integration, it is almost a straight line.
Expanding as a Taylor series around $x=\frac \pi 4$ (almost the middle of the interval), we have (I wrote a lot of terms knowing that most of them will not have any impact of the result)
$$\dfrac{1}{1+\left( \tan (x)\right) ^{\sqrt{2}}}=\frac{1}{2}-\frac{\sqrt{2}}{2}\left(x-\frac{\pi }{4}\right)+\frac{\sqrt{2}}{15} 
   \left(x-\frac{\pi }{4}\right)^5+\frac{16\sqrt{2}}{315}  \left(x-\frac{\pi
   }{4}\right)^7+\frac{13\sqrt{2}}{405}  \left(x-\frac{\pi }{4}\right)^9+$$ $$\frac{736
   \sqrt{2} }{31185}\left(x-\frac{\pi }{4}\right)^{11}+\frac{135098 \sqrt{2}}{6081075}\left(x-\frac{\pi }{4}\right)^{13}+O\left(\left(x-\frac{\pi}{4}\right)^{15}\right)$$
On the other side
$$\int _{\frac\pi 8}^{\frac\pi3} \left(x-\frac{\pi }{4}\right)^{2n+1}\, dx=\frac{ \left(\left(\frac{4}{9}\right)^{n+1}-1\right) \pi ^{2 n+2}}{2^{6 n+7}(n+1)}$$ Cumulating these terms and converting them to decimals, we have the following results
$$\left(
\begin{array}{c}
 0.3272492347 \\
 0.3575394439 \\
 0.3574868754 \\
 0.3574819953 \\
 0.3574816063 \\
 0.3574815691 \\
 0.3574815645 \\
 0.3574815638
\end{array}
\right)$$ while numerical integration would give $0.3574815637$.
We can have a &quot;nice looking&quot; approximation builging the $[4,4]$ Padé approximant of the integrand
$$\dfrac{1}{1+\left( \tan (x)\right) ^{\sqrt{2}}}\sim\frac{\frac 12 + \sum_{m=1}^4 a_m \left(x-\frac{\pi }{4}\right)^m}{1+ \sum_{m=1}^4 b_m \left(x-\frac{\pi }{4}\right)^m }$$ where the $a_m$ are
$$\left\{-\frac{\sqrt{2}}{2},-\frac{8}{21},\frac{8 \sqrt{2}}{21},\frac{1}{15}\right\}$$ and the $b_m$ are
$$\left\{0,-\frac{16}{21},0,\frac{2}{15}\right\}$$ Using partial fraction decomposition we end with
$$\int _{\frac\pi 8}^{\frac\pi3}\dfrac{dx}{1+\left( \tan (x)\right) ^{\sqrt{2}}}=\frac{5 \pi }{48}-\frac{173}{28} \sqrt{\frac{5}{13}} \tanh ^{-1}\left(\frac{40
   \sqrt{130} \pi ^2}{483840-4160 \pi ^2+7 \pi ^4}\right)-$$ $$\frac{10\sqrt{2}}{7} 
   \tanh ^{-1}\left(\frac{1600 \left(\pi ^2-273\right)}{651840-4160 \pi ^2+7 \pi
   ^4}\right)-\frac{40}{7} \sqrt{2} \coth ^{-1}(5)$$ which is $0.3574816071$.
As per the suggestion of @J.G the lower limit needs to be $\pi/6$, otherwise it is not doable by hand
Use the property that $$\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx.~~~~(1)$$
$$I=\int_{\pi/6}^{\pi/3} \frac{1}{1+\tan^{\sqrt{2}}x} dx~~~~(2)$$
Using (1), we get
$$I=\int_{\pi/6}^{\pi/3} \frac{1}{1+\tan^{\sqrt{2}}(\pi/2-x)} dx=\int_{\pi/6}^{\pi/3} \frac{1}{1+\cot^{\sqrt{2}}x} dx~~~~(2)$$
adding (2) and (3), we get
$$2I=\int_{\pi/6}^{\pi/3} \left(\frac{1}{1+\tan^{\sqrt{2}}x}+\frac{1}{1+\cot^{\sqrt{2}}x}\right)dx=\int_{\pi/6}^{\pi/3} 1 dx=\pi/6$$
$$\implies I=\pi/12$$