Question

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

Answer 1

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

Answer 2

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!

Answer 3

Thought I'd elaborate a bit on how I went from sin(x) to the rest of that:

if \[u = 4x+\pi, \ \implies x = \frac{u}{4}-\frac{\pi}{4}\]\[Then \sin(x) = \sin(\frac{u}{4} - \frac{\pi}{4}) = \sin(\frac{u}{4})*\cos(\frac{\pi}{4}) - \cos(\frac{u}{4})*\sin(\frac{\pi}{4}) = \frac{\sin(\frac{u}{4}) - \cos(\frac{u}{4})}{\sqrt{2}}.\]

Answer 4

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

Answer 5

Which is? I'm curious.

Answer 6

looks like almost the same trick as the other one.

Answer 7

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.

Answer 8

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.

Answer 9

\[math\neq \text{bag o' gimix}\]

Answer 10

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.

Answer 11

In a time limit or 3 hours*

Answer 12

I understand why there's a simpler solution, and it's convenient; but how would they disagree if it's a correct analytical method?

Answer 13

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.

Answer 14

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.

Answer 15

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.

Answer 16

I see what you mean. :P

Answer 17

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.

Answer 18

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!

Answer 19

Your arguments are basically the reason I attached it with the introduction. I guess we aren't disagreeing then.

Answer 20

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.

Answer 21

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.

Answer 22

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.

Answer 23

:D
PS if you like that sort of question here is another (see attached).

Answer 24

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

Answer 25

(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.

Answer 26

gotcha. thnx.

Answer 27

oh right! twice integrand = 1 gives \[\frac{\pi}{2}\] duh,got it.

Answer 28

wow I missed a lot of discussions here :)