# 4u integration qn (1 Viewer)

#### synthesisFR

##### Well-Known Member
lemme re try it one last time

#### Masaken

##### Unknown Member
guys I think I got it
can someone check tho it looks sus

#### howcanibesmarter

##### Well-Known Member
guys I think I got it
can someone check tho it looks sus
looks good to me

#### Trebla

Just a remark. Generally speaking, definite integrals where one of the points is undefined (x = 0 in this case) does not really appear in the HSC exam itself - if it ever does then they make it very clear you are taking the limiting value rather than evaluating it at that undefined end point.

#### tywebb

##### dangerman
Here is a quicker way:

$\bg_white \textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)\ \!dx=\frac{i}{2}[\text{Li}_2(e^{2ix})]_0^\pi=0\text{ where Li}_2\text{ is the dilogarithm function}$

$\bg_white \text{However}$

\bg_white \begin{aligned}\textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)dx&=\textstyle\int_0^\pi(\ln2+\ln(-i)+ix+\ln(\sin x))\ \!dx\\&=\textstyle\pi\ln2+\pi\cdot\frac{-\pi i}{2}+i\frac{\pi^2}{2}+\int_0^\pi\ln(\sin x)\ \!dx\\\textstyle\therefore \int_0^\pi\ln(\sin x)\ \!dx&=-\pi\ln2\end{aligned}

$\bg_white \textstyle\text{and by symmetry}\int_0^{\frac{\pi}{2}}\ln(\sin x)\ \!dx=-\frac{\pi}{2}\ln2$

#### synthesisFR

##### Well-Known Member
Here is a quicker way:

$\bg_white \textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)\ \!dx=\frac{i}{2}[\text{Li}_2(e^{2ix})]_0^\pi=0\text{ where Li}_2\text{ is the dilogarithm function}$

$\bg_white \text{However}$

\bg_white \begin{aligned}\textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)dx&=\textstyle\int_0^\pi(\ln2+\ln(-i)+ix+\ln(\sin x))\ \!dx\\&=\textstyle\pi\ln2+\pi\cdot\frac{-\pi i}{2}+i\frac{\pi^2}{2}+\int_0^\pi\ln(\sin x)\ \!dx\\\textstyle\therefore \int_0^\pi\ln(\sin x)\ \!dx&=-\pi\ln2\end{aligned}

$\bg_white \textstyle\text{and by symmetry}\int_0^{\frac{\pi}{2}}\ln(\sin x)\ \!dx=-\frac{\pi}{2}\ln2$
We are in high school

#### tywebb

##### dangerman
Here is a quicker way:

$\bg_white \textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)\ \!dx=\frac{i}{2}[\text{Li}_2(e^{2ix})]_0^\pi=0\text{ where Li}_2\text{ is the dilogarithm function}$

$\bg_white \text{However}$

\bg_white \begin{aligned}\textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)dx&=\textstyle\int_0^\pi(\ln2+\ln(-i)+ix+\ln(\sin x))\ \!dx\\&=\textstyle\pi\ln2+\pi\cdot\frac{-\pi i}{2}+i\frac{\pi^2}{2}+\int_0^\pi\ln(\sin x)\ \!dx\\\textstyle\therefore \int_0^\pi\ln(\sin x)\ \!dx&=-\pi\ln2\end{aligned}

$\bg_white \textstyle\text{and by symmetry}\int_0^{\frac{\pi}{2}}\ln(\sin x)\ \!dx=-\frac{\pi}{2}\ln2$
So there's this dude from Harvard University, Lars Ahlfors

He is like the coolest dude ever for complex numbers

- like the GOAT of complex numbers

who wrote the book Complex Analysis - probably the one which has been used more than any other for that subject.

My very short solution actually came from his book.

My version is somewhat less verbose than his, but essentially the same.

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#### AbdulkareemSyria

##### New Member
the Li refers to the polylogarithim function which is not assessed at all in school level maths

Edit: or the logarithmic integral which is the integral of 1/log x

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#### kendricklamarlover101

##### Member
surely if we use out of syllabus methods a competent examiner will give us full marks for it. like in this question for example if someone used tywebbs solution or even a contour integral we should be awarded full marks. iirc in last years ext 1 hsc people were still awarded marks for using the cross product rather than the intended method so whats the difference here?

#### H.S. Carslaw

##### New Member
in last years ext 1 hsc people were still awarded marks for using the cross product
Where did you get this information?

#### liamkk112

##### Well-Known Member
surely if we use out of syllabus methods a competent examiner will give us full marks for it. like in this question for example if someone used tywebbs solution or even a contour integral we should be awarded full marks. iirc in last years ext 1 hsc people were still awarded marks for using the cross product rather than the intended method so whats the difference here?
you’re kinda gambling tbh, in theory yeah you will be marked as correct however i’d advise to stay within the syllabus where possible. if you do use out of syllabus methods it would be a good idea to justify all the theorems you use to the best you can (which also is often much more tedious than just doing hsc methods), for example pretty sure people last year were marked down for using the cross product because they didn’t give enough reasoning, yeah it’s silly but it happens. whereas hsc methods are gonna be a safer bet, sometimes the longer way around but you’ll know for sure the marker cant mark u down for insufficient reasoning. eg if you’re using complex logs you’ll have to go through the whole reasoning as to how they work, what the integral is etc to be sure you’ll secure the marks. of course they may be lenient but there’s no telling and that’s kind of the issue

#### Trebla

surely if we use out of syllabus methods a competent examiner will give us full marks for it. like in this question for example if someone used tywebbs solution or even a contour integral we should be awarded full marks. iirc in last years ext 1 hsc people were still awarded marks for using the cross product rather than the intended method so whats the difference here?
I would like to see an official source comment on the use of out of syllabus methods. Intuitively, I struggle to believe there is much truth to the acceptance of such approaches as it creates all sorts of problems when trying to mark equitably. At best you likely need to derive these "out of syllabus" results from scratch using HSC syllabus methods before you utilise them.

For example, if you are asked to prove a 3 mark result that happens to be a rearrangement of a Taylor series then I don't think you deserve full marks by just quoting the Taylor series then rearrange it in one line, when the question intended for a proper derivation. There should be defined boundaries of what can be assumed and what needs to be mathematically argued in the question, otherwise it will be impossible to mark properly.

Also, practically speaking you cannot expect every HSC marker to be familiar with the infinitely many non-HSC concepts out there (they are maths educators, not mathematicians). So if your working is too obscure for them to understand then they might not accept it anyway, rightly or wrongly.

Basically agree with the above post, if you choose to use an out of syllabus method in the HSC exam (when it is perfectly doable by in syllabus methods) then you do so at your own risk...

#### tywebb

##### dangerman
Where did you get this information?
It came from senior marker in HSC Ext. 1 on HSC Feedback Day 2024. He said cross product method would be accepted because the question said "or otherwise". It's not really any more complicated than that.

#### iloveeggs

##### Active Member
oh my fuck what is this question

is this what im gonna be doing in the future i can't keep rawdogging math anymore omg

#### Average Boreduser

##### Rising Renewal
oh my fuck what is this question

is this what im gonna be doing in the future i can't keep rawdogging math anymore omg
its basically simultaneous equations they used here using a property lmfao. Don't be too worried about the looks. The complex number solution given here is out of syllabus.

#### Average Boreduser

##### Rising Renewal
oh my fuck what is this question

is this what im gonna be doing in the future i can't keep rawdogging math anymore omg
Also i'm curious as to where you learnt 'rawdog' from, because the way you used it seems rather/VERY concerning. (I'm assuming you may have learnt it from schoolmates without knowing its real connotations?)

#### iloveeggs

##### Active Member
Also i'm curious as to where you learnt 'rawdog' from, because the way you used it seems rather/VERY concerning. (I'm assuming you may have learnt it from schoolmates without knowing its real connotations?)
thanks, idk where i got it from but ik what it means lol

#### iloveeggs

##### Active Member
why does it feel like im in trouble now sorry i will mind my language this is a strictly educational forum frfr