It seems this case is always worked out by means of symmetry in standar Physics textbooks: arguing that for any two opposite segments in the ring, their perpendicular electric field components cancel each other. If the magnitude of the electric field of a small charge dq is:
$$dE=k\frac {dq}{x^2+a^2}$$
Then the $y$-component is:
$$dE_y=k\frac {dq}{x^2+a^2}\sin \theta$$
And the question is, how has this expression to be integrated in order to get zero?
If you would like to show without symmetry that the components go to zero, you need to write out the components using the unit-vector associated with the direction of $d\vec{E}$. This can be done by noticing that the vector connecting the element of charge and the point of interest also points in the same direction as $d\vec{E}$. Therefore, we can construct the unit-vector of $d\vec{E}$ from $\vec{r}$, the vector the points from the element of charge and the point of interest. The figure draw below is to allow you to see how I am generating the vector $\vec{r}$, which you can think of as the blue line.
So,
$$\vec{r}=\left<x-0,0-a\cos(\phi),0-a\sin(\phi)\right>=\left<x,-a\cos(\phi),-a\sin(\phi)\right>$$
The unit-vector is obtained by dividing by the magnitude of $\vec{r}$, which is $$r=\sqrt{x^2+(-a\cos(\phi))^2+(-a\sin(\phi))^2}=\sqrt{x^2+a^2}$$
So,
$$\hat{r}=\left<\frac{x}{x^2+a^2},-\frac{a\cos(\phi)}{x^2+a^2},-\frac{a\sin(\phi)}{x^2+a^2}\right>$$
Writing out $d\vec{E}$ fully, we get
$$d\vec{E}=k\frac{dq}{r^2}\hat{r}=k\frac{dq}{x^2+a^2}\left<\frac{x}{x^2+a^2},-\frac{a\cos(\phi)}{x^2+a^2},-\frac{a\sin(\phi)}{x^2+a^2}\right>$$
or
$$d\vec{E}=\left<k\frac{xdq}{\left(x^2+a^2\right)^{3/2}},-k\frac{adq\cos(\phi)}{\left(x^2+a^2\right)^{3/2}},-k\frac{adq\sin(\phi)}{\left(x^2+a^2\right)^{3/2}}\right>$$
Since the ring is a 1-dimensional object, we can associate it with
$$\lambda=\frac{dq}{dl}$$
or
$$dq=\lambda dl$$
And since the element $dl$ is just a section of the ring,
$$dl=ad\phi$$
The rest is just integrating the ring and you can see that the y-component and the z-component goes to zero because we are integrating $\sin(\phi)$ and $\cos(\phi)$ through a full period.
$$dE_{y}=k\frac{dq}{x^2+a^2}\sin\theta$$
Assume $\sigma$ is linear charge density.So, $dq=\sigma\ dx$
From the figure given,
$$\sin\theta=\frac{x}{(x^2+a^2)^{1/2}}$$
Putting the value of $\sin\theta$ and $dq$,
$$dE_{y}= k\sigma \frac{x\ dx}{(x^2+a^2)(x^2+a^2)^{1/2}}$$
As $y$-component is varying from $-a$ to $+a$,
so integrating it from $-a$ to $+a$, we get,
$$\int_{-a}^{+a}dE_{y}=k\sigma\int_{-a}^{+a}\frac{x\ dx}{(x^2+a^2)^{3/2}}$$
Take $x^2+a^2=t\implies2x\ dx=da$, and solving we get,
$$E_{y}=k\sigma\left[-\frac{1}{\sqrt{(x^2+a^2)}}\right]_{-a}^{+a}$$
So, $E_{y}= 0$
