Integral of arctangent of sine of cosine!

Substitute $t=x-\pi/2$ s.t integral is: $\Large\displaystyle\int_{-\pi/2}^{\pi/2}\large \underbrace{\arctan(\sin(\sin(x)))}_{\color{#D61F06}{\large\mathbf{Odd~ function}}} \mathrm{d}x$ $(\because \cos(x-\pi/2)=\sin x)$ $\Huge =\boxed 0$

(See here )

$I = \displaystyle \int_{0}^{\pi} \arctan(\sin(\cos(x))) dx$
Using,
$\displaystyle \int_{0}^{2a} f(x)dx = \int_{0}^{a} f(x) + f(2a-x)dx$
$I = \displaystyle \int_{0}^{\frac{\pi}{2}} \arctan(\sin(\cos(x)))dx + \arctan(\sin\cos(\pi-x)))dx = \int_{0}^{\frac{\pi}{2}} \arctan(\sin(\cos(x))) - \arctan(\sin(\cos(x)))dx = 0$

Let $I=\int_0^{\pi} \arctan(\sin(\cos(x))) dx$ using the property $\int_a^b f(x) dx= \int_a^b f(a+b-x) dx$ we have $I=\int_0^{\pi} \arctan(\sin(\cos(\pi-x))) dx=\int_0^{\pi} \arctan(\sin(-\cos(x))) dx=-I$ $I=-I\to I=0$

I was not happy when I ended up with this ugly integral after making the substitution $u= \cos x$ : $\large I=\int _{ -1 }^{ 1 }{ \cfrac { \arctan { \left( \sin { u } \right) } }{ \sqrt { 1-{ u }^{ 2 } } } } du$ However, I noticed that the integrand is an odd function , i.e. $f(-u)=-f(u)$ , therefore the integral of this function between $-1$ and $0$ is negative the integral between $0$ and $1$ . Thus the total integral equals $\large \color{#20A900}{\boxed{0}}$ .