91Ó°ÊÓ

Substitute \(\tan x\) with \(\cot (\pi/2 - x)\) in the integral \(\int_{0}^{\pi / 2} \frac{d x}{1+(\tan x)^{\sqrt{2}}}\). This gives us \(\int_{0}^{\pi / 2} \frac{d x}{1+(\cot (\pi/2 - x))^{\sqrt{2}}}\).

Perform the substitution \(x = \pi/2 - y\) and \(dx = -dy\) which now changes our integral to \(-\int_{\pi/2}^0 \frac{dy}{1+(\cot (y))^{\sqrt{2}}}\). If we swap the limits of integration, we can get rid of the negative sign. Thus our integral is \(\int_0^{\pi/2} \frac{dy}{1+(\cot (y))^{\sqrt{2}}}\). This integral is same to the original one, so if I is the original integral, we can write \(I = \int_{0}^{\pi / 2} \frac{d x}{1+(\tan x)^{\sqrt{2}}}\) = \(\int_0^{\pi/2} \frac{dy}{1+(\cot (y))^{\sqrt{2}}}\), it implies \(2I = \int_0^{\pi/2} \frac{dx}{1+(\tan (x))^{\sqrt{2}}} + \int_0^{\pi/2} \frac{dy}{1+(\cot (y))^{\sqrt{2}}} = \int_0^{\pi/2} \left[ \frac{1}{1+(\tan (x))^{\sqrt{2}}} + \frac{1}{1+(\cot (x))^{\sqrt{2}}}\right] dx\)

The integrand simplifies to \(\frac{1+(\cot (x))^{\sqrt{2}} + 1+(\tan (x))^{\sqrt{2}}}{1+(\tan (x))^{\sqrt{2}}+(cot (x))^{\sqrt{2}}}\) = \(\frac{(\sin^2 x + \cos^2 x) (\tan^{\sqrt{2}} x + \cot^{\sqrt{2}} x)}{(\tan^{\sqrt{2}} x)(\cot^{\sqrt{2}} x)}\) = \(1\). Therefore, \(2I = \int_0^{\pi/2} dx = [\frac{\pi }{2} - 0] = \frac{\pi }{2}\). Hence, \(I = \frac{\pi }{4}\)