Why Is Integration Difficult?

The Fundamental Theorem of Calculus^↗ says that integration and differentiation are the inverses of each other. So why is it that integration is extraordinarily, incomparably more difficult¹ than differentiation?

Well, this is true for a lot of things in life: often, reversing something is much, much harder than doing it in the first place – and in many cases, pretty much impossible. For instance, smash a fragile LEGO set, and it’ll take a while to repair it. It takes considerably more effort to rebuild it than it took to break it apart.² And if you’ve inevitably lost the instruction set, then it might not even turn out like the original.

A lot of things in maths are reversible. If we have a value $x$ and double it to give $2x$, we can perform the opposite of multiplication – division – to get our original value back.

If we plug a value into $e^x$, we can retrieve it with the inverse of exponentation, the logarithm.

But when we square a number, suddenly things aren’t so black-and-white. We might be inclined to say the “inverse” of squaring is the square root; however, when we have negative numbers that falls apart:

If they are inverses we would need $\sqrt{x^2}$ to give $x$. But as we know, $\sqrt{}$ only gives the positive root – which means $\sqrt{x^2} = |x|$ instead.

So why does that happen?

Think about what happens when we square $-2$. We have $(-2) \times (-2)$, so the twos multiply to give $4$. But the $-$ signs cancel out. It’s like they weren’t even there. In fact, just looking at the answer of $4$, there’s absolutely no way to tell the $-$ signs were there in the original calculation!

This is a loss of information. And unlike losing your password, it’s an unrecoverable loss.

We see the same thing in differentiation. When we have a constant term, like the $2$ in $x^2 + x + 2$, it vanishes when the expression is differentiated:

Integration is the inverse of differentiation, so integrating this expression should give us the original expression. But the $+2$ is missing, of course:

And this is why we need the $+c$ – it’s to account for any possible constant term.

In this case we knew $c = +2$, but there’s no way to know this when evaluating an integral blind.

Now this isn’t what makes integration difficult, but it helps illustrate the idea: when we differentiate, we often lose information. This information doesn’t have to be irreversibly lost, and it may not even be essential – but when we don’t have it, it can make a solution much tougher to spot.

This is perhaps my favourite example for demonstrating this. Let’s differentiate $\ln \left(\sec{x} + \tan{x} \right)$. We’ll use chain rule and quote the standard derivatives of logarithmic functions.

And miraculously, at the final step $\sec{x} + \tan{x}$ perfectly factors out, leaving us with just $\sec{x}$. Isn’t that beautiful. A rare instance of differentiation not being messy!

But now, suppose we do that in reverse. We want to integrate $\sec{x}$. And we have no clue of what we’ve just seen.

This is where it gets hard, because the critical factor of $\sec{x} + \tan{x}$ – it’s been lost. It was cancelled out in differentiation, just like $-$ when squaring. Without it, the integral is a whole lot more nontrivial, and the challenge lies in spotting that multiplication that allows you to apply [straight-up layer cake].

And so looking at this, one would wonder how you could ever spot the multiplication by $\sec{x} + \tan{x}$ when integrating $\sec{x}$.

This is an extreme case, but the same applies to many, many integrals.³

The challenge of integration is not in carrying out the integration,⁴ but in manipulating the integrand into a form where we can integrate it.

And how do we do that? Continue reading at How Do We Integrate?