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?
 A: 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.
A: Since the function is $2\pi$-periodic and continuous, you can change the integral to be on any interval of length $2\pi$:
$$\int_0^{2\pi} \tan(\cos(x))= \int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}} \tan(\cos(x)) dx = \int_{-\pi}^{\pi} \tan(\sin(x))dx = 0$$
where we used the substitution $x \mapsto \frac{\pi}{2}-x$ in the second step, yielding an integral of an odd function over a symmetric region.
A: $$ \int_0^{2\pi} \tan(\cos(x)) dx = \int_0^{\pi}\tan(\cos(x))dx + \int_{\pi}^{2\pi}\tan(\cos(x))dx$$
Subsitute in the latter $t = x- \pi$, so $dx = dt$ and $t \in (0,\pi)$, we get:
$$\int_0^{2\pi} \tan(\cos(x))dx = \int_0^\pi\tan(\cos(x))dx + \int_0^\pi \tan(\cos(t-\pi))dt$$
Since $\cos(a) = \cos(-a)$, we have $\cos(t-\pi) = \cos(\pi - t) = - \cos(-t) = -\cos(t)$
So we ended up (I'll just again swap that dummy variable to work with $x$ only)
$$\int_0^{2\pi} \tan(\cos(x)) dx = \int_0^\pi \tan(\cos(x)) + \tan(-\cos(x)) dx = \int_0^{\frac{\pi}{2}} \tan(\cos(x))dx + \int_0^{\frac{\pi}{2}}\tan(-\cos(x))dx + \int_{\frac{\pi}{2}}^\pi \tan(\cos(x))dx + \int_{\frac{\pi}{2}}^\pi \tan(-\cos(x))dx $$
Now, just to observe: $$ \int_0^{\frac{\pi}{2}} \tan(-\cos(x))dx = -\int_{\frac{\pi}{2}}^\pi \tan(\cos(x))dx $$
That can be seen substituting $t = \pi - x$ in the latter.
Similarly (with the same substitution), we have that: $$ \int_0^\frac{\pi}{2} \tan(\cos(x))dx = - \int_{\frac{\pi}{2}}^\pi \tan(-\cos(x))dx$$
So we're just ended up with something like $a + (-b) + b + (-a) = 0 $, where
$$a = \int_0 ^\frac{\pi}{2} \tan(\cos(x))dx , b = \int_\frac{\pi}{2}^\pi \tan(\cos(x))dx $$
A: Let $t=\cos x
  \implies \sqrt {1-t^2}dt=dx$
Limits change from $0$ to $2\pi $
To $1$ to $1$
Limit $1$ to $1$ is obviously $0$
A: 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: $$I=\int_{0}^{2\pi}\tan(\cos(x))dx=-\int_{-\pi}^{\pi} \tan(\cos(x))dx=0$$
