I like integrals 3

Write it as:- $\large\mathfrak G= \displaystyle\int_{-\pi/2}^{\pi /2} \dfrac{\frac ab \sec^2 x~\mathrm{d}x}{ \left(\frac{a^2}{b^2}\tan^2 x+1\right)^{3/2}}$ Substitute $\frac ab\tan x=\tan y$ such that $\frac ab \sec^2 x~\mathrm{d}x=\sec^2 y~\mathrm{d}y$ .

$\large\therefore\mathfrak G=\displaystyle\int_{-\pi/2}^{\pi /2} \dfrac{ \sec^2 y~\mathrm{d}y}{ \underbrace{\left(\tan^2 y+1\right)^{3/2}}_{\sec^3 y}}$ $\large =\displaystyle\int_{-\pi/2}^{\pi /2}\cos y~\mathrm{d}y$ $\large =2\displaystyle\int_0^{\pi/2}\cos y~\mathrm{d}y=\Large\boxed{\color{#007fff}{2}}$