Evaluate ∫limits_0^(π)/2 1/(1 + (tan x)^2/3) dx.

The correct answer is Option (4) → $\frac{\pi}{4}$

$I = \int\limits_{0}^{\frac{\pi}{2}} \frac{1}{1 + (\tan x)^{2/3}} dx \dots(i)$

$I = \int\limits_{0}^{\frac{\pi}{2}} \frac{1}{1 + \left[ \tan \left( \frac{\pi}{2} - x \right) \right]^{2/3}} dx$

[Using property $\int_{0}^{a} f(x) dx = \int\limits_{0}^{a} f(a-x) dx$]

$I = \int\limits_{0}^{\frac{\pi}{2}} \frac{1}{1 + (\cot x)^{2/3}} dx$

$I = \int\limits_{0}^{\frac{\pi}{2}} \frac{(\tan x)^{2/3}}{(\tan x)^{2/3} + 1} dx$

$I = \int\limits_{0}^{\frac{\pi}{2}} \frac{(\tan x)^{2/3} + 1 - 1}{(\tan x)^{2/3} + 1} dx$

$I = \int\limits_{0}^{\frac{\pi}{2}} \frac{1 + (\tan x)^{2/3}}{1 + (\tan x)^{2/3}} dx - \int\limits_{0}^{\frac{\pi}{2}} \frac{1}{1 + (\tan x)^{2/3}} dx$

$I = \int\limits_{0}^{\frac{\pi}{2}} 1.dx - I \text{ [From eq. (i)]}$

$2I = \int\limits_{0}^{\frac{\pi}{2}} 1.dx$

$2I = [x]_0^{\pi/2}$

$2I = \frac{\pi}{2}$

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