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watchmath

  • 5 years ago

Compute \[\int_0^{\pi/2}\frac{\sin x}{\sin(x+\frac{\pi}{4}}\]

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  1. watchmath
    • 5 years ago
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    \[\int_0^{\pi/2}\frac{\sin x}{\sin(x+\frac{\pi}{4})}\,dx\]

  2. anonymous
    • 5 years ago
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    This one's a pain in the retrice :P You've got, after simplifying, \[\int\limits_{0}^{\pi/2}{\csc(x+\pi/4)\sin(x)}dx\] and, for the moment ignoring the limits of the integral,\[u = 4x + \pi, du = 4 \ dx\] \[\implies \frac{-1}{4} \ \int\limits \ {\frac{\csc(\frac{u}{4})(\sin(\frac{u}{4})-\cos(\frac{u}{4}))}{\sqrt{2}}} du\] because of the compound angle formula, and then you can simplify it to \[\frac{-1}{4\sqrt{2}} \ \int\limits \ {1-\cot(\frac{u}{4})} \ du\] and I bet you know how to proceed after that, with the limits. Hope I helped!

  3. anonymous
    • 5 years ago
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    Thought I'd elaborate a bit on how I went from sin(x) to the rest of that: if \[u = 4x+\pi, \ \implies x = \frac{u}{4}-\frac{\pi}{4}\]\[Then \sin(x) = \sin(\frac{u}{4} - \frac{\pi}{4}) = \sin(\frac{u}{4})*\cos(\frac{\pi}{4}) - \cos(\frac{u}{4})*\sin(\frac{\pi}{4}) = \frac{\sin(\frac{u}{4}) - \cos(\frac{u}{4})}{\sqrt{2}}.\]

  4. anonymous
    • 5 years ago
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    Although your method is pretty impressive, it's actually much nicer if you DON'T ignore the limits; there is a 'trick'.

  5. anonymous
    • 5 years ago
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    Which is? I'm curious.

  6. anonymous
    • 5 years ago
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    looks like almost the same trick as the other one.

  7. anonymous
    • 5 years ago
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    It's actually pretty cheeky just asking this question on its own as it is part of a larger one in which you prove the result you need to use first and then apply it. But you should expand the denominator and then u = pi/2 - x. See what comes out.

  8. anonymous
    • 5 years ago
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    You should try that first, but here is the full question (see attached). Note: I am not doing this to ruin the question (you have already solved it); besides, I feel justified because a) I sent him the questions earlier and b) there are other integrals, some of which use the same trick and the other a variation, which are also fun.

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  9. anonymous
    • 5 years ago
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    \[math\neq \text{bag o' gimix}\]

  10. anonymous
    • 5 years ago
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    That question is (one of the easiest ever) questions for entrance to Cambridge Mathematics, the best University in the world (to put it in context, you would need to answer 4-6 questions equal or harder to that [generally much harder]). So I think they would disagree with your assessment of this question.

  11. anonymous
    • 5 years ago
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    In a time limit or 3 hours*

  12. anonymous
    • 5 years ago
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    I understand why there's a simpler solution, and it's convenient; but how would they disagree if it's a correct analytical method?

  13. anonymous
    • 5 years ago
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    not that i object to these puzzle, i like them. but they are just that: puzzles. if you figure out the key, you can open the door. that is fine. no problem. the fact that this is on an entrance exam does not give it some authority. it is a fine question. no argument. but if you don't know the gimmick you will probably not figure it out on the fly.

  14. anonymous
    • 5 years ago
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    I know that for most formal evaluations even if you use a correct method that doesn't conform to the way it's conventionally solved, you usually get full marks.

  15. anonymous
    • 5 years ago
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    I didn't say they disagree. Your method (also Wolfram Alpha's) would do fine, but I thought you'd like to see the 'intended' one. Perhaps it's more relevant in, say, the second question in that example (and the last) which CANNOT be worked out indefinitely in standard terms.

  16. anonymous
    • 5 years ago
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    I see what you mean. :P

  17. anonymous
    • 5 years ago
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    Hence why I said it was bad to ask it alone, without the starting part, satellite. I would not expect many people to spot the trick.

  18. anonymous
    • 5 years ago
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    this is exactly why every day math students, the ones who are not going to MIT or Cambridge (again, not that there is anything wrong with that!) think math is incomprehensible gibberish taught by evil wizards intent on trickery. oh look, i know a trick and you don't!

  19. anonymous
    • 5 years ago
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    Your arguments are basically the reason I attached it with the introduction. I guess we aren't disagreeing then.

  20. anonymous
    • 5 years ago
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    i repeat i happen to like these. they are challenging and fun to work on. what makes math interesting. but to give them to students as an entrance exam smacks of mathematics at it worst.

  21. anonymous
    • 5 years ago
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    Have you looked at my attachment? In case you haven't, note that it is NOT just asked as in the original post. The 'trick' is given to you and you have to prove it works, THEN apply it. I AGREE posting it by itself here was a bad idea.

  22. anonymous
    • 5 years ago
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    oh posting i here is fine as a challenge. and if the trick is given to you to prove and then asked to be applied to this specific example, that is find too. so i should just be quiet. what gets my goat, because i see it frequently, is asking a question that relies on some trick (ok mathematical fact) that they probably do not know. here as a post it is fine. and on an exam it is fine too with the proviso that you are given the relevant facts in advance. so i guess i agree with you as well, if it is in that context.

  23. anonymous
    • 5 years ago
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    :D PS if you like that sort of question here is another (see attached).

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  24. anonymous
    • 5 years ago
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    when i sober up i will look at it. also figure out why it is obvious that \[\int_0^{\frac{\pi}{2}}\frac{sin^n(x)}{cos^n(x)+sin^n(x)}=\frac{\pi}{4} \]

  25. anonymous
    • 5 years ago
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    (I wouldn't say it is obvious), but for the same reason the integral in this thread worked (let u = pi/2 - x, then add the result to the original integrand for a nice total of 1.

  26. anonymous
    • 5 years ago
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    gotcha. thnx.

  27. anonymous
    • 5 years ago
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    oh right! twice integrand = 1 gives \[\frac{\pi}{2}\] duh,got it.

  28. watchmath
    • 5 years ago
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    wow I missed a lot of discussions here :)

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