Using symmetry

$\begin{aligned} I & = \int_0^\pi \cos (5x) \cos (7x)\cos(11x) \cos (19x)\cos(27x)\ dx \quad \quad \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x)\ dx \\ & = \frac 12 \int_0^\pi \left(\cos (5x) \cos (7x)\cos(11x) \cos (19x)\cos(27x)+\cos (5\pi-5x) \cos (7\pi - 7x)\cos(11\pi - 11x) \cos (19\pi-19x)\cos(27\pi -27x)\right)\ dx \\ & = \frac 12 \int_0^\pi \left(\cos (5x) \cos (7x)\cos(11x) \cos (19x)\cos(27x)+\cos (\pi-5x) \cos (\pi - 7x)\cos(\pi - 11x) \cos (\pi-19x)\cos(\pi -27x)\right)\ dx \\ & = \frac 12 \int_0^\pi \left(\cos (5x) \cos (7x)\cos(11x) \cos (19x)\cos(27x)-\cos (5x) \cos (7x)\cos(11x) \cos (19x)\cos(27x)\right)\ dx \\ & = \frac 12 \int_0^\pi 0\ dx \\ & = \boxed{0} \end{aligned}$

As the problem suggests this problem is about using symmetry. We could use the product of cosines rule to write the integral in terms of a sum cosines, integrate to get a sum sines, and then evaluate to get 0, but exploiting symmetry gets us the answer much faster. $cos(x)$ is an odd function about $\pi/2$ . That is to say $\cos(\pi/2+x)=-\cos(\pi/2-x)$ . $\cos(nx)$ is odd about $\pi/2$ , when n is odd and even when n is even. $\int_0^\pi \cos(nx)$ where n is odd is 0, because $\cos(nx)$ is odd about $\pi/2$ and we are integrating equally far on both sides of $\pi/2$ . $\int_0^\pi \cos(nx)$ where n is even is also 0. When two odd functions are multiplied together the result is an even function. When an even function is multiplied by an odd function the result is an odd function. Thus, when three odd functions are multiplied together the result is an odd function. When multiplying odd functions together the result will switch back and forth between odd and even functions. When an even number of odd functions are multiplied together the result is even and when an odd number of odd functions are multiplied together the result is an odd function. Thus we are integrating an odd function equally distant from the point around which it is odd. Thus, the answer is 0.

Another way to see the symmetry is to convert the problem to an integral of a product of sines. Since $\cos(x)=-\sin(x-\pi/2)$

$\displaystyle\int_{0}^{\pi} \cos(5x)\cos(7x)\cos(11x)\cos(19x)\cos(27x) dx=-\int_{0}^{\pi} \sin(5x-\pi/2)\sin(7x-\pi/2)\sin(11x-\pi/2)\sin(19x-\pi/2)\sin(27x-\pi/2) dx$

We can use the u substitution $u=x-\pi/2$ to get

$\displaystyle-\int_{-\pi/2}^{\pi/2} \sin(5u)\sin(7u)\sin(11u)\sin(19u)\sin(27u) du=0$

In this case the symmetry is clearer as it is around u=0.