Answer: Integral of 1/(1+tan^√2 x)

Key Idea / Intuition

The exponent $\sqrt{2}$ looks frightening, but it's a red herring. The trick is to pair the integral with itself under the substitution $x \mapsto \frac{\pi}{2} - x$, which swaps $\tan$ and $\cot$. The two integrals add to give exactly $\pi/2$, so each equals $\pi/4$ — regardless of the exponent.

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Formal Proof / Solution

Let $\alpha = \sqrt{2}$ (the argument works for any $\alpha > 0$). Define

$I = \int_0^{\pi/2} \frac{1}{1 + \tan^\alpha x}\, dx.$

Step 1: Apply the substitution $x \mapsto \frac{\pi}{2} - x$.

Under this substitution $dx \mapsto -dx$, and the limits swap (then flip back), giving:

$I = \int_0^{\pi/2} \frac{1}{1 + \tan^\alpha\!\left(\frac{\pi}{2}-x\right)}\, dx.$

Step 2: Use the identity $\tan\!\left(\frac{\pi}{2} - x\right) = \cot x = \frac{1}{\tan x}$.

$I = \int_0^{\pi/2} \frac{1}{1 + \cot^\alpha x}\, dx = \int_0^{\pi/2} \frac{1}{1 + \tan^{-\alpha} x}\, dx = \int_0^{\pi/2} \frac{\tan^\alpha x}{1 + \tan^\alpha x}\, dx.$

Step 3: Add the two expressions for $I$.

$2I = \int_0^{\pi/2} \frac{1}{1+\tan^\alpha x}\, dx + \int_0^{\pi/2} \frac{\tan^\alpha x}{1+\tan^\alpha x}\, dx = \int_0^{\pi/2} \frac{1 + \tan^\alpha x}{1 + \tan^\alpha x}\, dx = \int_0^{\pi/2} 1\, dx = \frac{\pi}{2}.$

Conclusion:

$\boxed{I = \frac{\pi}{4}}$

The value is completely independent of the exponent $\alpha$ — so $\sqrt{2}$ was there purely as a distraction.