Integration - U substitution

Can someone help me? I learned Integration by substitution and I just don't like the method. I don't like how it abuses the $\frac{dy}{dx} $ notation. I always understood that $\frac{dy}{dx} $ couldn't be treated as a fraction, since it's not a fraction.

Would there be a way to integrate functions of the form $f(g(x))$ without using $u$-substitution?

Also, could someone explain how $u$-substitution even works to get the correct answer?

Answer

You can rigorously justify $u$-substitution by using the fundamental theorem of calculus and the chain rule. But before doing this, lets look at a concrete example. Take the following integral:

$$

\int_0^1 \frac{2x}{x^2 + 1}dx
$$

Examine the integrand and notice that $\frac{d}{dx}(x^2 + 1) = 2x$. That is, the derivative of the denominator is equal to the numerator. This is the type of setup that you look for when using substitution. We let $u = x^2 + 1 \implies du = 2xdx$. Substituting these in to the integral, and changing the limits of integration to $u(0)$ and $u(1)$ we get:

$$
\int_1^2 \frac{1}{u}du
$$

And so we have the result:

$$
I_1 = \int_0^1 \frac{2x}{x^2 + 1}dx = \int_1^2 \frac{1}{u}du
$$

Now, let's justify this. Let $f$ be a real valued function continuous on the interval $[a,b]$. Let $g$ be a real valued differentiable function on $[a,b]$. Consider the following integral:

$$
I_2 = \int_a^b f(g(u))g'(u)du
$$

This is analogous to the first integral I wrote with $f(x) = \frac{1}{x}$ and $g(x) = x^2 + 1$. Let $F$ be an antiderivative of $f$. Then we have:

$$
\frac{d}{du}F(g(u)) = F'(g(u))g'(u) = f(g(u))g'(u)
$$

So $F(g(u))$ is an antiderivative of $f(g(u))g'(u)$ and by the fundamental theorem of calculus we have:

$$

I_2 = F(g(b)) - F(g(a))
$$

But also we can say that

$$
F(g(b)) - F(g(a)) = \int_{g(a)}^{g(b)}f(t)dt
$$

And so we have shown that:

$$
I_2 = \int_a^b f(g(u))g'(u)du = \int_{g(a)}^{g(b)}f(t)dt
$$

Notice that $I_1$ is directly analogous to $I_2$.