Written by Colin+ in integration.

Some time ago, I recommended the mnemonic “LIATE” for integration by parts. Since you have a choice of which thing to integrate and which to differentiate, it makes little sense to pick something that’s hard to integrate as the thing to integrate.

With that in mind, you would look down the list:

- Logarithms
- Inverse trigonometric functions
- Algebraic functions
- Trigonometric functions
- Exponentials

… and pick whichever showed up first as the thing to differentiate.

However, a commenter noted that they had learnt the rule as ILATE, with the first pair switched. Which is better? There’s only one way to find… with an **integration-off**!

Bring on the first (and indeed only) challenge: given that $I = \int \ln(x) \arccos(x) \dx$, find $I$.

According to LIATE, the way to integrate this would be to take $\ln(x)$ as the bit to differentiate – which I’ll call $u$, as per convention, and $\arccos(x)$ as the bit to integrate – which I’ll call $v’$.^{1}

$u=\ln(x)$ is easy to differentiate: you end up with $u’=\frac{1}{x}$.

$v’ = \arccos(x)$ is much harder to integrate. The conventional method is to do it by parts; with $U=\arccos(x)$ and $V’=1$ to get $v=x \arccos(x) – \sqrt{1-x^2}$.

Putting it together using the parts formula $I = uv – \int u’ v dx$, we get

$I=\ln(x)\br{x \arccos(x) – \sqrt{1-x^2}} – \int \br{\arccos(x) – \frac{\sqrt{1-x^2}}{x}}\dx$

This took me a lot longer than it should have done to work out^{2} This was mainly due to tiredness; a method is to substitute $u^2 = 1-x^2$ and work with partial fractions on the result. It turns out to be $\sqrt{1-x^2}-\ln(\sqrt{1-x^2}+1)+\ln(x)$. Plus a constant, if that’s your sort of thing.

We can integrate $\arccos(x)$ and we can integrate $\frac{\sqrt{1-x^2}}{x}$, so we’re done: $I = \br{\ln(x)-1}\br{x \arccos(x)-\sqrt{1-x^2}} + \sqrt{1-x^2}-\ln(\sqrt{1-x^2}+1)+\ln(x)$. Plus a constant. That’s a mess, but it agrees with WA. Phew.

This time, with $u = \arccos(x)$, we get $u’ = -\frac{1}{\sqrt{1-x^2}}$; with $v’ = \ln(x)$, we get $v = x\br{\ln(x)-1}$, using a similar trick as for arccosine.

The integral via parts is $I = x\br{\ln(x)-1}\arccos(x) + \int \frac{x\br{\ln(x)-1}}{\sqrt{1-x^2}} \dx$

Now we can apply parts again, with $U=\ln(x)-1$ and $V’ = \frac{x}{\sqrt{1-x^2}}$, so that $U’=\frac{1}{x}$ and $V = -\sqrt{1-x^2}$.

This gives $(1-\ln(x))\sqrt{1-x^2}+ \int \frac{\sqrt{1-x^2}}{x}dx$

And that final integral is the same as the one from the aside above.

Putting it all together gives $I = x\br{\ln(x)-1}\arccos(x) + (1-\ln(x))\sqrt{1-x^2} + \sqrt{1-x^2}-\ln(\sqrt{1-x^2}+1)+\ln(x)$. Plus a constant – again, agreeing with WA.

In honesty, neither of the two alternatives stands out as a great deal worse than the other. Both involve some pretty sharp integration, and I’d be inclined to put the two on the same footing at the top.

The moral of the story: mnemonics aren’t set in stone; I can imagine situations where either of the factors may be preferable as a thing to differentiate or to integrate. ILATE and LIATE are useful guides, but developing your instinct for how to approach nasty integrals is much better than slavishly following them.