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?
Best Answer
The integrand is bounded between $\pm\tan 1$, so the integral converges. Since $\tan\theta$ is odd, $\tan\cos(\pi-x)=-\tan\cos x$. Thus $\int_0^\pi\tan(\cos x)dx=0$. The $\int_\pi^{2\pi}$ part follows similarly.