Let $A$ be the arc of the unit circle in the first quadrant,
that is, the quarter-circle arc between $(1,0)$ and $(0,1)$.
Then integrating along the arc using $ds$ to denote
the length measure, the length of the arc is
$$
\int_A ds = \frac\pi2.
$$
Now let's parameterize the path along the arc by some variable $t$
via a one-to-one continuous mapping $f$ from the interval $[0,1]$ to the
points on the arc with $f(0) = (1,0)$, so that
$$
\int_0^1 \frac{ds}{dt} \, dt = \int_A ds = \frac\pi2.
$$
Notice that with this parameterization, $\frac{dx}{dt} \geq 0$
and $\frac{dy}{dt} \leq 0$ everywhere on the arc.
At any point along the arc,
$\frac{ds}{dt}$ is the hypotenuse of a right triangle
with legs $\frac{dx}{dt}$ and $-\frac{dy}{dt}$.
Because the direction of the path at each point $(x,y)$ is perpendicular
to the segment from $(0,0)$ to $(x,y)$,
the triangle with sides $\frac{ds}{dt}$,
$\frac{dx}{dt}$, and $-\frac{dy}{dt}$
is similar to a right triangle with hypotenuse $1$ and legs $x$ and $y$.
Hence
$$
\frac{\frac{dx}{dt} - \frac{dy}{dt}}{x + y}
= \frac{\left(\frac{ds}{dt}\right)}{1} = \frac{ds}{dt}.
$$
As $t$ runs from $0$ to $1$, $x$ runs from $0$ to $1$ and
$y$ runs from $1$ to $0$. Therefore
\begin{align}
\int_0^1 \frac{dx}{x + y} + \int_0^1 \frac{dy}{x + y} &=
\int_0^1 \frac{dx}{x + y} - \int_1^0 \frac{dy}{x + y} \\
&= \int_0^1 \frac{\frac{dx}{dt}\,dt}{x + y}
- \int_0^1 \frac{\frac{dy}{dt}\,dt}{x + y} \\
&= \int_0^1 \frac{\frac{dx}{dt} - \frac{dy}{dt}}{x + y}\,dt \\
&= \int_0^1 \frac{ds}{dt}\,dt \\
&= \frac\pi2.
\end{align}
Then you can use the observation that
$$\int_0^1 \frac{dx}{x + y} = \int_0^1 \frac{dy}{x + y}$$
to conclude that $$\int_0^1 \frac{dx}{x + y} = \frac\pi4.$$
The trick here, as in the parameterization $x=\cos\theta$,
$y=\sin\theta$, is that you have to "match up" the meshes of
$dx$ and $dy$ so that you can correctly add the them together.
The trigonometric substitution was just one particular mapping
from a parameter $\theta$ to the arc.
But you don't need to invoke that particular mapping.
The mapping $x = t$, $y = \sqrt{1 - t^2}$ also works, for example.
Update: The following is an observation on notation
inspired by the comments.
The solution above is "dressed up" (or you might say "dumbed down")
to fit with my recollection of what a student would become accustomed to
in a first-year calculus course based on standard analysis.
The parameter $t$ sets a direction of integration along the arc
and keeps the symbols $dx$, $dy$, and $ds$ from having to appear outside the usual contexts such as $\int (\text{something})\,dx$
or $\frac{dx}{d(\text{something})}$,
but it really has no other reason to be defined.
Given a decent theory of differentials or infinitesimals,
so that we can speak of $ds$, $dx$, and $dy$ as objects in their own rights,
if we orient the arc so that $\int_A$ integrates from $(0,1)$ to $(1,0)$
then we have the ratios $ds : dx : -dy = 1 : y : x$ and can write
\begin{align}
\frac{dx - dy}{x + y} &= \frac{ds}{1} = ds, \\
\int_0^1 \frac{dx}{x + y} + \int_0^1 \frac{dy}{x + y}
&= \int_0^1 \frac{dx}{x + y} - \int_1^0 \frac{dy}{x + y} \\
&= \int_A \frac{dx}{x + y} - \int_A \frac{dy}{x + y} \\
&= \int_A \frac{dx - dy}{x + y} \\
&= \int_A ds \\
&= \frac\pi2.
\end{align}
This is a more intuitive approach; or at least, I think it is, since it is
how I got the intuition on how to set up the integrals in the first place,
before I added the parameterization.
And now for a third version inspired by additional comments!
Under the orientation of $ds$ along the quarter-circle arc $A$
from $(1,0)$ to $(0,1)$, we have
$\frac{dx}{ds} = y$ and $\frac{dy}{ds} = -x$. Then
\begin{align}
\int_0^1 \frac{dx}{x + y} + \int_0^1 \frac{dy}{x + y}
&= \int_0^1 \frac{dx}{x + y} - \int_1^0 \frac{dy}{x + y} \\
&= \int_A \frac{dx/ds}{x + y}\,ds - \int_A \frac{dy/ds}{x + y}\,ds \\
&= \int_A \frac{y - (-x)}{x + y}\,ds \\
&= \int_A ds \\
&= \frac\pi2.
\end{align}
