Replace x with π/2 - x to get the equivalent integral
[tex]\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx[/tex]
but the integrand is even, so this is really just
[tex]\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx[/tex]
Substitute x = 1/2 arccot(u/2), which transforms the integral to
[tex]\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du[/tex]
There are lots of ways to compute this. What I did was to consider the complex contour integral
[tex]\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz[/tex]
where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be
[tex]\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}[/tex]
which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit
[tex]\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}[/tex]
and it follows that
[tex]\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}[/tex]
