It looks like the integral is
[tex]\displaystyle 6 \int yz\cos(x) \, ds[/tex]
With [tex]x=t[/tex], [tex]y=\cos(t)[/tex], and [tex]z=\sin(t)[/tex], the line element is
[tex]ds = \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2 + \left(\dfrac{dz}{dt}\right)^2} \, dt = \sqrt2 \, dt[/tex]
and so
[tex]\displaystyle 6 \int yz\cos(x) \, ds = 6\sqrt2 \int_0^\pi \cos^2(t) \sin(t) \, dt[/tex]
Substitute [tex]u = \cos(t)[/tex] and [tex]du=-\sin(t)\,dt[/tex], then
[tex]\displaystyle 6 \int yz\cos(x) \, ds = -6\sqrt2 \int_1^{-1} u^2 \, du = 12\sqrt2 \int_0^1 u^2 \, du = \boxed{4\sqrt2}[/tex]
where in the second-to-last equality, we have
[tex]\displaystyle -\int_1^{-1} = \int_{-1}^1[/tex]
and
[tex]\displaystyle \int_{-1}^1 u^2 \, du = 2 \int_0^1 u^2\,du[/tex]
by symmetry.
