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Edexcel A2 C4 Mathematics June 2016 - Official Thread

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Original post by BBeyond
Fair probs should have guessed :tongue: Camb?


He's not that good. :wink:
These problems are getting somewhat above c4 standard..:tongue:
Original post by 1 8 13 20 42
These problems are getting somewhat above c4 standard..:tongue:


:teehee:
Original post by Zacken
3. Using θ↦π4−θ\theta \mapsto \frac{\pi}{4} - \theta, evaluate: ∫0π/4ln⁡(1+tan⁡θ) dθ\displaystyle \int_0^{\pi/4} \ln (1 + \tan \theta) \, \mathrm{d}\theta.

Extension: Evaluate ∫01ln⁡(1+x)1+x2 dx\displaystyle \int_0^{1} \frac{\ln (1 + x)}{1+x^2} \, \mathrm{d}x.


Original post by Zacken
4. Evaluate ∫0π/2ln⁡sin⁡θ dθ\displaystyle \int_0^{\pi/2} \ln \sin \theta \, \mathrm{d}\theta using a similar substitution to my other integral.

5. Generalise this substitution and prove that it holds for all smoothly behaved ff when evaluating ∫abf(x) dx\displaystyle \int_a^b f(x) \, \mathrm{d}x


Saving these beauties for later :smile:
Original post by Euclidean
Saving these beauties for later :smile:


Let me know how you get on. :yep:
Original post by Zacken
4. Evaluate ∫0π/2ln⁡sin⁡θ dθ\displaystyle \int_0^{\pi/2} \ln \sin \theta \, \mathrm{d}\theta using a similar substitution to my other integral.

5. Generalise this substitution and prove that it holds for all smoothly behaved ff when evaluating ∫abf(x) dx\displaystyle \int_a^b f(x) \, \mathrm{d}x


Ffs misread your first question here and have been trying to integrate ln(1+sinx) instead and getting absolutely nowhere :colonhash:
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Original post by Zacken
If I say x↦f(x)x \mapsto f(x), I'm really just making a substitution u=f(x)u = f(x) then u=xu=x. :-)

Well if you call your integral II - then you have I=ln⁡2−II = \ln 2 - I. :wink:

(although that's not quite correct, use that idea).


:redface: I've never seen that before. I still don't know what to do though. If I let the original integral = I and mine = J, I have that J = the integral of ln(2) - I? Is that the right idea? :smile:
Original post by Zacken
:teehee:


I just hope if any non-STEPers/maths students/general maths enthusiasts see them they don't get too scared..

I'll throw in a standard that everyone should know to level the balance

∫1+cosxdx\displaystyle \int \sqrt{1 + cosx} dx
Original post by Fudge2
:redface: I've never seen that before. I still don't know what to do though. If I let the original integral = I and mine = J, I have that J = the integral of ln(2) - I? Is that the right idea? :smile:


Don't worry about not seeing it before, it's a little advanced. :redface:

If we let the original integral be I=∫0π/4ln⁡(1+tan⁡θ) dθI = \int_0^{\pi/4} \ln (1 + \tan \theta) \, \mathrm{d}\theta, then our substitution gives us:

I=∫0π/4ln⁡2−ln⁡(1+tan⁡θ) dθ=∫0π/4ln⁡2 dθ−∫0π/4ln⁡(1+tan⁡θ) dθ\displaystyle I = \int_0^{\pi/4} \ln 2 - \ln (1 + \tan\theta) \, \mathrm{d}\theta = \int_0^{\pi/4} \ln 2 \, \mathrm{d}\theta - \int_0^{\pi/4} \ln(1+\tan\theta) \, \mathrm{d}\theta

Which is just: I=∫0π/4ln⁡2 dθ−II = \int_0^{\pi/4} \ln 2 \, \mathrm{d}\theta - I, so you can re-arrange and solve for II. :smile:
(edited 8 years ago)
Original post by BBeyond
Ffs misread your first question here and have been trying to integrate ln(1+sinx) instead and getting absolutely nowhere :colonhash:


Yeah, the mis-read one is a hard one. The definite integral is given in terms of Catalan's constant: 2C−π2ln⁡22\mathcal{C} - \frac{\pi}{2}\ln 2. :teehee:
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Original post by Zacken
Don't worry about not seeing it before, it's a little advanced. :redface:

If we let the original integral be I=∫0π/4ln⁡(1+tan⁡θ) dθI = \int_0^{\pi/4} \ln (1 + \tan \theta) \, \mathrm{d}\theta, then our substitution gives us:

I=∫0π/4ln⁡2−ln⁡(1+tan⁡θ) dθ=∫0π/4ln⁡2 dθ−∫0π/4ln⁡(1+tan⁡θ) dθ\displaystyle I = \int_0^{\pi/4} \ln 2 - \ln (1 + \tan\theta) \, \mathrm{d}\theta = \int_0^{\pi/4} \ln 2 \, \mathrm{d}\theta - \int_0^{\pi/4} \ln(1+\tan\theta) \, \mathrm{d}\theta

Which is just: I=∫0π/4ln⁡2 dθ−II = \int_0^{\pi/4} \ln 2 \, \mathrm{d}\theta - I, so you can re-arrange and solve for II. :smile:


Haha I was kinda hoping that was the case! That's really neat though. I got pi/8ln2...?
Original post by Fudge2
Haha I was kinda hoping that was the case! That's really neat though. I got pi/8ln2...?


Yeah, nothing like this would ever come up on a C4 paper. That's correct, good work!
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Original post by Zacken
Yeah, nothing like this would ever come up on a C4 paper. That's correct, good work!


Wooo! Thanks for your guidance :biggrin:
Original post by Fudge2
Wooo! Thanks for your guidance :biggrin:


Enjoyed it? :biggrin:
Original post by Zacken
4. Evaluate ∫0π/2ln⁡sin⁡θ dθ\displaystyle \int_0^{\pi/2} \ln \sin \theta \, \mathrm{d}\theta using a similar substitution to my other integral.

5. Generalise this substitution and prove that it holds for all smoothly behaved ff when evaluating ∫abf(x) dx\displaystyle \int_a^b f(x) \, \mathrm{d}x

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Original post by Zacken
Enjoyed it? :biggrin:


Yeah! Despite the fact I'll probably never have to use that trick again it was cool...:biggrin:
Original post by 1 8 13 20 42
∫1+cosxdx\displaystyle \int \sqrt{1 + cosx} dx


22sinx2+C2\sqrt{2}sin\frac{x}{2} + C

Spoiler

(edited 8 years ago)
Original post by BBeyond


Yep, that's fine. But unwieldy but it works. Now try doing the general case. :biggrin:
Original post by Fudge2
Yeah! Despite the fact I'll probably never have to use that trick again it was cool...:biggrin:


I've used it a hundred times over when doing integration problems, not at C4 level, but you'll see. :-)
Original post by Zacken
Yep, that's fine. But unwieldy but it works. Now try doing the general case. :biggrin:


Is there a quicker way of doing that? Buzzing with that though ahah didn't even know about this trick until today. I think that might be a bit past my level lol

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