Good luck solving this INSANE INTEGRAL | Tricky A-Level Maths Question #52
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Sorry everyone! At 2:21 I overlooked parts as a possible solution method. You can solve with parts using the recursive method.
Can you please make a video on how to do it am extremely struggling on it
@@mardinjordan8642 Don't worry about it too much this is an extreme example
you can do it by parts, it gives that the initial integral is x(3^lnx) -(ln3)(the integral) you just add it to the other side and then divide by 1+ln3
Learning maths from you from Sydney Australia 🇦🇺 Great work!
adw1z's simplification is best, but requires some down-field vision. the usub in the video is the best obvious approach, but e^u*3^u you want to convert the 3 to e^(log3), and then you can combine. technically you get another u-type substitution, but a linear one so you can just "reverse chain rule" it
beautiful question
Thanks
A very quick trick is to notice: 3^ln(x) = exp(ln3)^ln(x) = exp(ln(x))^ln(3) = x^(ln3), which integrates to:
(ln(3) + 1)^-1 x^(ln(3) + 1) + C
= x* 3^ln(x) / (ln(3) + 1) + C
Edit: I just read now, u need a ln(x) sub, which is very annoying. But it makes life much easier using the above method still: u = lnx so e^u = x, e^u du = dx, and then integrate e^(uln(3))e^u du = e^(u(1+ln3)) du, which becomes:
1/(1+ln3) e^(u(1+ln3)) + C
= x* 3^ln(x) / (ln(3) + 1) + C
Avoids any sort of parts or partial differentiation for 7 marks :)
Yep, did it this way pretty quickly.
That’s genius 👏🏿
Very good but the question asks us to use the substitution u = lnx. If we didn't do that would we be marked down?
@@spencergee6948 I wouldn't risk it, so I'd try and use a u = ln(x) sub wherever possible. Even in the shortcut method u can use it, which saves a lot of time and then u are guaranteed all the marks
i got up to (3e)^u and got ln(3e) * (3e)^u. from there i ended up with 3^(ln(x)) * x(1 + ln(3)) + c, but the 1+ln(3) is meant to be the denominator so i just divided everything by (1 + ln(3))^2 and got (3^(ln(x)))/(1 + ln(3)) + c/((1 + ln(3))^2) and since c/((1+ln(3))^2) is just a constant i just renamed it and turned it into + C on its own. that got me the thing u had to show. would that work or is that the wrong method?
Isn’t 3^lnx, equal to (e^ln3)^lnx,
= (e^lnx)^ln3
= x^ln3
So does it not just integrate to
(x^(ln3 + 1))/(ln3 + 1)
Genius 😳
That's correct but the question said u need to use the substitution method
the better method is that x^ln3 = 3^lnx. So its mere instant, obviously its not allowed in the q but yeah.
u can still cheese it by doing the u = ln(x) sub, x^(ln3) dx = e^(u(1 + ln(3))) du, its just as easy and saves lots of time
recursive method is perfect for this
Why can't we use e = pi = 3 ? Or is integration by engineers not allowed?
Sir at 4:42 why did you consider to differentiate 1/lna. Where did this came from?
I think it's because integration and differentiation are inverse functions, and when you integrate a function you multiply by 1 over the function. Not entirely sure though.
Easy🤣love how easy high school is compared to stuff later down the road
where do you find these questions? I saw some people mention madasmaths but I can't find these questions. Can you please send me the link?
I was astounded by your knowledge in the missing xi shorts, so here’s a harder one: Which club links:
-Dean Moxey
-Adam Randell
-Mike Williamson
That’s mighty tough! I don’t even know who the 2 top two are. I know Williamson played for Newcastle so I’ll guess them!
@@MrAstburyMaths Dean Moxey played at Crystal Palace while Randell plays for Plymouth after a loan spell at this club. Mike Williamson came through their academy and Neil Warnock now acts as an advisor. It was Torquay united, tough one to be fair, had to research a couple of them
By parts work well
yeah you jus gotta rearrange the parts
@@Fate212 fr why is bro tripping its easy
fr its quicker than this way
Yes sorry my mistake, parts does work well here
Sorry for tripping, my mistake, parts does work well here
hi could anyone please explain how to do this question by parts?
u=e^u, dv/dx=3^u substitute into parts formula
du/dx=e^u, v=3^uln3
I havent tried this but maybe 3^lnx x one, choose one as the value to integrate. Kinda similar to how when yoi integrate lnx you integrate lnx x 1.
For parts this is how I did it:
So if you sub u = lnx you get the integral of 3^u e^u as shown in the video
Now parts:
z = 3^u so dz/du = ln(3)*3^u (ofc u normally say u = something then du/dx = something but in this case we have already used the letter u so that would make things confusing)
dv/du = e^u so v = e^u
integral of e^u*3^u = e^u*3^u - integral of ln(3)*3^u*e^u (Using by parts)
Now notice that when u do it by parts you have the same integral again. So u can say I = integral of e^u*3^u
So:
I = e^u*3^u -ln(3)*I
Hence I = e^u*3^u/(ln3+1)
Then sub back in u = lnx, then u get the thing they want, let me know if i explained something badly
@@ronakkumar3295 Thank you very much I did it again and now it works!
@@ronakkumar3295 hey the first part is all good but i dont understand how the( ln3 I )became (ln3 + 1) ? and the denominator
I solved this integral with parts and sub very easily the easiest integral ever why did u say parts wont get u far??
u integrate 3^u and differentiate e^u then add the integral to the other side and simplify so easy?
Hi sorry you found the integral too easy. It’s been hard trying to find super hard questions everyday and record, edit and post them. Yes, you are right parts does work well, sorry for my mistake there. I think both methods are equally acceptable and efficient.
@@MrAstburyMaths its alr G, ill forgive ur incompetence just this time (￣y▽,￣)╭
how did you integrate the integral when doing it by parts
@@jamzfym u need to treat it like an equation and "add or subtract" the rhs to lhs then simplfy
@@Mathiscool-jx9mvwhy did you make the u= e^u
i cant believe i actually managed to do it first try tbf it just all flowed well but I'm most worried about the modelling quadratic questions that come up ahha I never know how to do them even though they're meant to be the easier marks
bro no the exponential modelling qs shake me to my core
Nice work 👌
where is this question from? Im trying to find out how to do this by parts thanks 😃
Madas SYN Paper W
How did you know to use e^u?
After differentiating, try sub in x*du = dx, you will get get 3^u * x du which has the integral in terms of u and x, so you would need to get x in terms of u, which would be e^u. Sir here has just instantly put it to e^u before x as usually these types of problems require you to do that, but not all of course.
May i ask, how do you know that by part swill be trickier?
Sorry my mistake, parts does work well here.
You can’t do this by parts, I tried it and it puts you into an infinite loop- to show what I mean, Try using parts on ∫3^u e^u du , you should find that within the very first part, the repeated integral step in the formula is the exact same; hence, an infinite loop
@@trqpurpz7402 You need to know the recursive step for integration by parts in order for it to work. This type of step is not on the Edexcel Spec. That's why I didn't think about it during the video. It does work though!
@@trqpurpz7402 do it once then add it on i think
@@MrAstburyMaths wow, maths is so interesting! I just watched a recursive integration step video and I understand it. If I was to use this in the exam, would I lose marks is my question?
whats wrong with saying xdu=dx?
because you don't want any "x" terms in your integral when you're using u-substitution- you just want it in terms of "u" so you have to convert all "x" terms to "u" terms
May I ask what exam board this question is from?
Edexcel
@@julesbourlot5684 international or normal?
this isnt off any of the actual edexcel papers so far this is from a MADASMATHS synoptic paper ( W I think ) and this paper is a question of a paper rated above the a level standard difficulty
@@ZeroWinz Ok thank you so much mate.
Use reduction formula
i think this question is harder than alevel
lightwork 😀
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