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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?

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    $\begingroup$ Try splitting into $(0,\pi)$ and $(\pi,2\pi)$ then reflect each interval $\endgroup$ – Edward H. Sep 8 '19 at 20:01
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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.

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    $\begingroup$ Actually you don't even need $\int_\pi^{2\pi}$, but anyway, nice insight! (+1) $\endgroup$ – Mostafa Ayaz Sep 8 '19 at 20:02
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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.

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$$ \int_0^{2\pi} \tan(\cos(x)) dx = \int_0^{\pi}\tan(\cos(x))dx + \int_{\pi}^{2\pi}\tan(\cos(x))dx$$

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

$$a = \int_0 ^\frac{\pi}{2} \tan(\cos(x))dx , b = \int_\frac{\pi}{2}^\pi \tan(\cos(x))dx $$

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