Integral with Tricks

Similar solution with D G 's

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac 1{1+\tan^{\sqrt 3}x} dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac 1{1+\tan^{\sqrt 3}x} + \frac 1{1+\tan^{\sqrt 3}\left(\frac \pi 2 - x\right)} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac 1{1+\tan^{\sqrt 3}x} + \frac 1{1+\cot^{\sqrt 3}x} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac 1{1+\tan^{\sqrt 3}x} + \frac 1{1+\frac 1{\tan^{\sqrt 3}x}} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac 1{1+\tan^{\sqrt 3}x} + \frac {\tan^{\sqrt 3}x}{\tan^{\sqrt 3}x + 1} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 dx = \frac \pi 4 \end{aligned}$

Therefore, $A = \boxed{4}$ .

$G = \int_0^{\pi/2} \frac{1}{1 + \tan^a{x}} dx = \int_0^{\pi/2} \frac{\cos^a{x}}{\cos^a{x} + \sin^a{x}} dx$

Using the fact that $\int_{\alpha}^{\beta} f(x) dx = \int_{\alpha}^{\beta} f(\alpha + \beta - x) dx$ :

$G = \int_0^{\pi/2} \frac{\cos^a{(\pi/2 - x)}}{\cos^a{(\pi/2 - x)} + \sin^a{(\pi/2 - x)}} dx = \int_0^{\pi/2} \frac{\sin^a{x}}{\sin^a{x} + \cos^a{x}} dx$ $2 G = \int_0^{\pi/2} \frac{\sin^a{x} + \cos^a{x}}{\sin^a{x} + \cos^a{x}} dx = \int_0^{\pi/2} dx = \frac{\pi}{2}$ $G = \frac{\pi}{4}$