SciForums.com > Science > Physics & Math > sine^n(x) PDA View Full Version : sine^n(x) Post ReplyCreate New Thread John Connellan03-08-04, 01:48 PMCan some one please tell me what is: ∫sinnx I can't be bothered working it out with integration by parts! P.s. can someone tell me how to get math html symbols up too!!! mathman03-08-04, 05:47 PMsinx=(exp(ix)-exp(-ix))/2i. Use the binomial theorem to expand sin^n(x) as a sum of terms of the form cexp(ikx) (c depends on k) k will take on values from n to -n in steps of 2. You can then integrate term by term. Continue this type of manipulation in reverse (exp back to sin or cos) and get a finite trig series. Dinosaur03-09-04, 10:04 AMJohn Connellan: Try using <> instead of [] for subscripts & superscripts. AD103-09-04, 10:31 AMThe special character codes appear as follows e.g. For ∫, the code is & int ; with no spaces. You can find the integral ∫ sinn(x) dx without complex numbers by using reduction formulae. John Connellan03-10-04, 06:10 AMIs this right: ∫sin2x = 1/2(x)-1/4 (sin 2x) + C and ∫sin3x = -1/3(sin2x)(cos x) -2/3(cos x) + C Dinosaur03-10-04, 01:32 PM&int Above Used & int without the space. It did not show as an integral sign. AD103-10-04, 05:50 PMDinosaur, you didn't put the semicolon on the end. John Connellan03-11-04, 01:35 PM&int Above Used & int without the space. It did not show as an integral sign. AD1 told me to do it without spaces! Anyway here we go again. ∫sin(x) I hope that worked! What I was actually asking though is if the result of the integration above was right?! John Connellan03-11-04, 01:40 PMIm gonna make this my integration 'questioning' thread if that oks. Can some one tell me what the answer to this is? ∫sin((x-1)*c) AD103-11-04, 02:00 PMDistribute the c within the brackets: ∫ sin((x - 1)c) dx = ∫ sin(cx - c) dx The general form is ∫ sin(ax + b) dx = -(1/a)cos(ax + b) + C The phase angle is irrelevant, only the coefficient of x is relevant. So in this case, the answer is -(1/c)cos(cx - c) + C John Connellan03-11-04, 03:14 PMThanks a lot AD1. I should really start reading up on some calculus! I only know the basics. James R03-12-04, 02:53 AMTo do: Integral [sin ((x-1)c)] dx, first substitute u = c(x-1) Then du/dx = c or, in other words dx = (1/c) du. Therefore Integral [sin(c(x-1))] dx = Integral [sin u (1/c)] du = -(1/c) cos u = -(1/c) cos (c(x-1)) + constant John Connellan03-15-04, 02:00 PMHow would u go about integrating this (2-wave) function? ∫ A1Sin(wx+b)+A2Sin(hx+d)+c dx AD103-15-04, 02:09 PMUse the method given above and integrate each term seperately. Your integral is then: -(A1/w) cos(wx + b) - (A2/h) cos(hx + d) + x + C. John Connellan03-15-04, 02:19 PMOK sorry. Knew how to do each part seperately but forgot that in integration u CAN just do them seperately when addition is involved :rolleyes: ryans03-15-04, 06:20 PMJohn Connellan I can't be bothered working it out with integration by parts! I'll give you a hint if you want to be a successful physicist or mathematician. BE BOTHERED. You sound as if you treat these problems as a chore, you're in a rush to solve them. But I bet you can sit for hours on the internet and read a whole lot of infodribble. Trust me, there is no greater satisfaction in arriving as some formula or method that someone else did 50, 100 etc years ago. Also by doing these problems you will learn little tricks which will help you with harder problems in the future. If you are going to be an engineer or a chemist, ignore that and ask someone for the answer, it makes us feel important lethe03-15-04, 09:19 PMBut I bet you can sit for hours on the internet and read a whole lot of infodribble. infodribble. hahaha. that is the best description i have ever heard of the kind of stuff we get on sciforums. i like that term. Trust me, there is no greater satisfaction in arriving as some formula or method that someone else did 50, 100 etc years ago. except perhaps arriving at some formula that no one has ever arrived at before ryans03-15-04, 10:37 PMYes, but I usually make so many assumptions on the way that it is totally irrelevant to any problem John Connellan03-16-04, 06:04 AMYou sound as if you treat these problems as a chore, you're in a rush to solve them. Thats true actually. Only because I read a book on calculus before and was quickly bored once I got beyond the basics. Im beginning to realise what ur talking about now though so I want to start learning again. Post ReplyCreate New Thread