STEP Tricks and Hacks Thread

just a few suggestions, you might want to add in what's wrong with the AM-GM inequality "proof" that you showed and how to fix it

P.S: Nice touch adding the proof by contrapositive, I just want to add my thoughts on it for anybody who doesn't see why $A \implies B \iff \neg B \implies \neg A$

Will edit now.

I'm a big fan of the contrapositive

One recurring theme is to exploit the symmetry of trigonometric functions during integrating. Using this in conjunction with:



is particularly useful.

Here's an example: $\int_0^{\pi/4} \log (1 + \tan \theta) \, \mathrm{d}\theta$. How is this relevant here? Call our integral $I$, then using the substitution $\theta \mapsto \frac{\pi}{4} - \theta$ we have:



Extension to this, using the same idea, roughly - attempt $\displaystyle \int_0^{\pi/2} \log \sin \theta \, \mathrm{d}\theta$

Am I on the right track?




Posted from TSR Mobile

You messed up your negative signs, you should get $\int \ln \sin \theta \, \mathrm{d}\theta = \int \ln \cos \theta \, \mathrm{d}\theta \neq -\int \ln \cos \theta \, \mathrm{d}\theta$ with the limits from $0\,$ to $\pi/2$ on each integral but I can't be arsed typesetting.

Then you should add both integrals to get $2I = \int \ln (\sin \theta \cos \theta) \, \mathrm{d}\theta$ with the usual limits and then think about $\sin 2\theta$.

But doesn't the substitution change the limits from 0 to pi/2, to pi/2 to 0 so then I switch it by adding a negative sign.

I actually did it that way the first time I tried the integral but then I noticed that the substitution changes the limits.




Posted from TSR Mobile

Yes - but you forgot that $u = \pi/2 - x$ adds in a factor of $-1$ since $\mathrm{d}u/\mathrm{d}x = -1$.

Oh **** yes. Also, I remember that then you get an integral with ln 2 or something like that as integrand which is easy to calculate but the other integral with an integrand $\sin2\theta$ and then you do the same addition of 1/2 integral of cos + 1/2 integral of sin but with $2\theta$ this time and then you continue ad infinitum. How does this simplify it? I can't find the workings out on this integral I did yesterday, so I might be wrong.

$\displaystyle 2I = \int_0^{\pi/2} \ln \left(\frac{1}{2} \sin 2\theta \right) \, \mathrm{d}\theta = \int_0^{\pi/2} \ln \frac{1}{2} \, \mathrm{d}\theta + \int_0^{\pi/2} \ln \sin 2\theta \, \mathrm{d}\theta$

But you want an $I$ in there somewhere, so the sub $u = 2x$ ($x \mapsto \frac{x}{2}$) gets you that and some nice thinking about the symmetry of sine gets you the home run.

I got $I = -\frac{\pi}{3}\ln 2$

Haha, you're the third person to give me that incorrect answer since I've posted that problem.

We know that $A \implies B$ is a true statement literally every single time unless we have the case that $A$ is false and $B$ is true, we can rewrite this as: $A \implies B$ is a true statement unless $\neg A \wedge B$.

The limits change after the sub $u = 2x$.

Look at some truth tables. https://en.wikipedia.org/wiki/Truth_table

The conditional connective, is false, if, and only if the premise is true but the conclusion is false.

And I did change them? Then I split the integral into limits consisting of $0 \text{ to } \pi/2$ each.

Oh, yes.

You seem to be saying that $\int_0^{\pi} \ln \sin u \, \mathrm{d}u = 2 \times \frac{1}{2} \int_0^{\pi/2} \ln \sin u \, \mathrm{d}u$.

You want to say that $\int_0^{\pi} \ln \sin u \, \mathrm{d}u = 2I \neq I$

$u = 2\theta \Rightarrow \frac{du}{d\theta} = 2 \text{, so } d\theta = \frac{1}{2} du$ which is where this 1/2 comes from.

But you multiplied the entire thing by $2$ the line before.

We have $2I = -\frac{\pi}{2} \ln 2 + \frac{1}{2} \int_0^{\pi} \ln \sin u \ \mathrm{d}u$

You decided to multiply everything by $2$ to get: $4I = -\pi \ln 2 + \int_0^{\pi} \ln \sin u \, \mathrm{d}u$.

But we know that the last term is $2I$ so $4I = -\pi \ln 2 + 2I \iff I = -\frac{\pi}{2}\ln 2$

Oh yes, I saw my mistake. I think that was more of a silly mistake to be quite honest.

This was a really fun question, thanks a lot.