This approach is the hardest for visualizing vectors!!!
Assuming a uniform charge distribution about the ring, the linear charge density lambda will be Q / (pi a), where we are assigning the variable a to be the radius of the ring. In terms of the charge density, the infinitesimal charge element will be dq = lambda ds = lambda a d(theta).
phi is also introduced now to get ready for consideration of components of dE. Note that phi is constant once the point of evaluation is chosen; it is does not vary as the angle theta is varied as the charge distribution is spanned.
Begin by focusing on the left half of the ring. The infinitesimal contribution dE will point directly away from the charge infinitesimal, along the line connecting it and the point of evaluation.
The symmetry is exploited by considering the four conjugate elements shown.
The vertical and axial components of dE from one charge infinitesimal will always be balanced by the contribution from its conjugate.
Thus, only the horizontal x-components of the four dE need be integrated. All four quadrants of charge contribute the same horizontal component, so we need only integrate from 0 to pi / 2 and multiply the answer by 4.
It takes some maneuvering to visualize exactly what component of dE is needed, but the right perspective shows that it is proportional to sin(theta) and sin(phi).