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?

  • 1
    $\begingroup$ It may comfort you to know that the abuse of notation is simply a way to remember how the substitution works. Substitution can be rigorously justified, but this is typically not shown in a first calculus course. $\endgroup$ – wgrenard Nov 17 '16 at 19:01
  • $\begingroup$ Would the justification not be accessible for a student with knowledge up to first year calc? +wgrenard $\endgroup$ – adsadasd Nov 17 '16 at 19:02
  • $\begingroup$ It is accessible, though can be complicated to look at at first if you aren't already comfortable with using u substitution. See my answer. $\endgroup$ – wgrenard Nov 17 '16 at 19:45

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 $I_2$.


It probably depends on the form of $f(g(x))$. However, substitution works like this,

$I = \int f(g(x))dx$

$ u = g(x) \Rightarrow du = g'(x)dx \Rightarrow dx = \frac{1}{g'(x)}du$

$I = \int f(u)\frac{1}{g'(g^{-1}(u))} du$

Note that the limit changes too. For example, if limits are from $x=a$ to $x=b$ then new limits become $u=g(a)$ to $u=g(b)$. However, you can always integrate with respect to $u$, then again substitute $u=g(x)$ and use limits of $x$.


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