Cut it into dq, then keep the direction

The charge element is written with the density that matches the geometry:

$$dq = \lambda\,dx \ \text{(line)}, \qquad dq = \sigma\,dA \ \text{(surface)}, \qquad dq = \rho\,dV \ \text{(volume)}.$$

Each element contributes a field at the point of interest with magnitude $dE = k\,dq/r^2$, directed along the line from THAT element to the field point. The setup that survives contact with a real problem:

1. Draw one general element, somewhere in the middle of the distribution, and label its position with the integration variable.
2. Write $r$ for THAT element, in terms of the integration variable. This is the step people skip.
3. Write $dE$, then resolve it into components using the angle that element makes.
4. Decide which components survive by symmetry, and integrate only those.
5. Check limits: far away the answer must collapse to $kQ/r^2$.

Worked case, a rod along its own axis. Rod of length $L$ and total charge $Q$ lies from $x = d$ to $x = d + L$, and the field point is at the origin. Here $\lambda = Q/L$ and every element pushes along the same axis, so no components are needed:

$$E = \int_{d}^{d+L} \frac{k\lambda\,dx}{x^2} = k\lambda\left[\frac{1}{d} - \frac{1}{d+L}\right] = \frac{kQ}{d(d+L)}.$$

Test it: for $d \gg L$ the denominator becomes $d^2$ and the result is $kQ/d^2$, as it must be. For $d$ comparable to $L$ it is nothing like $kQ/d^2$, which is the entire reason the integral exists.

Constants come out. Anything that depends on the integration variable stays in. The trap is that for one famous geometry, the ring on its axis, almost everything is constant, and students carry that convenience into geometries where it is false.

Ring on its axis. Radius $R$, total charge $Q$, field point on the axis a distance $z$ from the center. Every element is the same distance $\sqrt{z^2 + R^2}$ away and makes the same angle with the axis, so both $r$ and $\cos\theta = z/\sqrt{z^2+R^2}$ genuinely come out:

$$E = \frac{k\cos\theta}{z^2+R^2}\int dq = \frac{kQz}{(z^2+R^2)^{3/2}}.$$

At $z = 0$ this gives zero, and for $z \gg R$ it becomes $kQ/z^2$. Both checks pass.

Rod on its perpendicular bisector. Now nothing comes out. An element at position $x$ along the rod sits a distance $r = \sqrt{x^2 + y^2}$ from the field point, and its angle changes with $x$ too, so the projection factor is $y/\sqrt{x^2+y^2}$ and it also stays inside:

$$E = \int_{-L/2}^{L/2} \frac{k\lambda\,dx}{x^2+y^2}\cdot\frac{y}{\sqrt{x^2+y^2}} = \frac{kQ}{y\sqrt{y^2 + L^2/4}}.$$

Limits again: $y \gg L$ gives $kQ/y^2$, and $L \gg y$ gives $2k\lambda/y$, the infinite line. Two independent checks on one formula, and both are cheap.

Writing $E = (k/r^2)\int dq = kQ/r^2$ for this geometry is the point-charge shortcut smuggled back in through the integral sign.

On the perpendicular bisector of a uniform rod, pair each element on the left with its mirror image on the right. Their components ALONG the rod are equal and opposite and cancel; their components along the bisector are equal and add. So the net field points along the bisector, perpendicular to the rod, and only that integral has to be done.

Two ways to get this wrong, and they are opposites.

1. Symmetry ignored. Grinding through both integrals, including the one that is zero, and making a sign error in it. If the mirror argument holds, say so and drop that component before integrating.
2. Symmetry invented. Waving at symmetry at a point that is NOT on the bisector, near one end of the rod for instance. There is no mirror image there, both components survive, and both integrals are real work.

State the mirror explicitly before you use it: for every element at $+x$ there is an identical element at $-x$, equidistant from my field point. If you cannot say that sentence about your geometry, the cancellation is not available.

The same argument runs on an arc. For a semicircular arc of radius $R$ and total charge $Q$, every element is exactly $R$ from the center, so $r$ comes out. The angle does NOT, because each element points a different way. Projecting onto the symmetry axis and integrating $\sin$ over the half-circle gives

$$E_{\text{center}} = \frac{2k\lambda}{R} = \frac{2kQ}{\pi R^2},$$

using $\lambda = Q/(\pi R)$. Note $2/\pi \approx 0.64$, so the true field is about a third smaller than the naive $kQ/R^2$: that missing third IS the projection.

Ask what direction the answer should have BEFORE integrating, and the arithmetic gets a free check. For a symmetric distribution and a field point on its symmetry axis, the net field lies along that axis:

1. Uniform rod, point on the perpendicular bisector: field is perpendicular to the rod, along the bisector, pointing away from a positively charged rod.
2. Ring, point on the central axis: field is along the axis, never tangent to the ring and never radial in the ring's plane.
3. Semicircular arc, point at the center: field is along the arc's symmetry axis, pointing away from the arc for positive charge.

Reporting a field parallel to the rod, or tangent to the ring, means the transverse components were never cancelled and the parallel ones never were either. A resultant cannot point along the direction whose contributions cancel in pairs.

And the converse matters just as much. Move the field point off the axis and the symmetry is gone: near one end of a rod the field tilts, with a component along the rod that no pairing removes. The direction is then something you compute, not something you assert.