Integrate With Inverse

Relevant wiki: Integration Tricks

First note: $\small{\color{#69047E}{\tan^{-1} \left( e^{\sin x} \right) =\cot^{-1}\left( e^{-\sin x} \right) =\dfrac{\pi}2-\tan^{-1} \left( e^{-\sin x} \right)}}$ Next, denote the integral by $\mathfrak R$ and apply $\displaystyle\int_a^bf(x)\mathrm{d}x=\int_a^bf(a+b-x)\mathrm{d}x$ and add the two to get:

$2\mathfrak R=\displaystyle\int_{-2}^2(\tan^{-1} \left( e^{-\sin x} \right) +\color{#69047E}{\tan^{-1} \left( e^{\sin x} \right) })\mathrm{d}x$

$=\displaystyle\int_{-2}^2(\cancel{\tan^{-1} \left( e^{-\sin x} \right) }+\color{#69047E}{\dfrac{\pi}2\cancel{-\tan^{-1} \left( e^{-\sin x} \right)}})\mathrm{d}x$

$\large =\displaystyle\int_{-2}^{2}\dfrac{\pi}2\mathrm{d}x=2\pi$

$\Large\therefore\mathfrak{R}=\pi\approx\boxed{3.14}$

Using properties of Definite Integrals, $\displaystyle \int_{-a}^a f(x)\, dx = \int_0^a f(x) + f(-x)\, dx$

So $\begin{aligned}I &= \int_0^2 \arctan \left(e^{\sin x}\right)+\arctan \left(e^{\sin (-x)}\right)\,dx\\ &= \int_0^2 \arctan \left(e^{\sin x}\right)+\arctan \left(\dfrac1{e^{\sin x}}\right)\,dx\end{aligned}$

Now since $e^{\sin x} >0$ for all real values of $x$ , $\arctan \dfrac1x = \mathrm{arccot }~x = \dfrac\pi2 - \arctan x$

So, $I$ becomes $\begin{aligned}I &= \int_0^2 \arctan \left(e^{\sin x}\right)+\dfrac\pi2 - \arctan \left(e^{\sin x}\right)\,dx\\ &= \int_0^2 \dfrac\pi2\,dx\\ \Large&= \Large\boxed{\Large\pi}\end{aligned}$