Interesting integral $\int_0^{2\pi}\tan(\cos(x))dx=0$ Consider the integral  $$I=\int_0^{2\pi}\tan(\cos(x))dx$$
I would like to show this integral is $0$ via elementary methods (excluding complex analysis, special functions, series representations). 
The bounds of integration suggest some kind of symmetry argument to show that the integral vanishes. 
I tried $x=\pi/2-u\implies dx=-du\implies$ $$I=-\int^{-\frac{3\pi}{2}}_{\frac{\pi}{2}}\tan(\sin(u))du$$ From here I don’t see a good route.
I also tried $$I=\int_0^{2\pi}\tan(\cos(x))dx=\int_0^{2\pi}\frac{\sin(\cos(x))}{\cos(\cos(x))}dx$$ Then let $t=\cos(\cos(x))\implies dt=-\sin(\cos(x))\cdot-\sin(x)=\sin(x)\sin(\cos(x)) \space dx$
Now $$I=\int_0^{2\pi}\frac{\csc(x)}{t}dt$$
The question now would be how to invert $t=\cos(\cos(x))$? But this obviously would be tough. Again, I think there’s a simple symmetry argument I’m missing. Can anyone help?

We can deal with this kind of integral question by switching to the variable: $y=\pi+x$.$$I=\int_{0}^{2\pi} \tan(\cos(x))dx=\int_{-\pi}^{\pi} \tan(\cos(\pi+x))dx=-\int_{-\pi}^{\pi}\tan(\cos(x))dx$$
Antiderivatives of odd functions are even and vice versa. Since tangent function is odd, its antiderivative
is even. This boils down to the fact that: $$I=\int_{0}^{2\pi}\tan(\cos(x))dx=-\int_{-\pi}^{\pi} \tan(\cos(x))dx=0$$
