Consider the basic integral, $\int \frac{dx}{1+x^2}$. The ''standard'' substitution is $u = \tan(x)$ (why?). However, the way I've always done these things is look at a triangle with sides $1,x,\sqrt{1+x^2}$ with angle $\theta$, and from there reasoned to the point that $\int \frac{dx}{1+x^2} = \int d\theta = \theta + C = \arctan(x) + C$ with the dubious $dx = \sec^2(\theta)d\theta$.
While this is obviously better than relying on ''standard'' substitutions from an intuitive perspective, is there any way to make this entirely rigorous without any hand-waving, relying only on the fundamental theorem of calculus? I'm struggling to think how exactly I should format the argument.
Let $f(x) = \frac{1}{1+x^2}$. We will look at the right triangle with angle $x$ and sides $1,u(x),\sqrt{1+u(x)^2}$. Here, $u = \tan(x)$ and its derivative, $u'(x) = \sec^2(u(x))$. We also note from the triangle, that $\frac{1}{1+x^2} = \cos^2(u(x)).$
Now we're ready to integrate: $f(u(x))u'(x) = \frac{\sec^2(u(x))}{\sec^2(u(x))} = 1$. By the fundamental theorem of calculus, $$\int f(u(x))u'(x)dx = \int 1\,dx = x + C = F(u(x)) + C\text{, where $C \in \mathbb{R}$.} $$
Now we know that $x = \arctan(u(x))$, so the integral evaluates to $F(u(x)) = \arctan(u(x)) + C$. Finally, because $\tan(x)$ defines a homeomorphism $(-\pi/2,\pi/2)\to\mathbb{R}$, $F(x) = \arctan(x) + C$.
Additionally, we can also do this substitution the other way around. Looking at the right triangle with angle $u$ and sides $1,x,\sqrt{1+x^2}$, we get that $x(u) = \tan(u)$, and $x'(u) = \sec^2(u)$.
Then we get $\int f(x(u))x'(u)\,du = \int 1\,du = u + C = \arctan(x(u)) + C = F(x(u)) + C.$