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 2011-04-10, 13:44 #1 Baztardo   Apr 2011 38 Posts Need help with an integral Hi there! New to this forum and not good at math.. I wonder if anyone of you could help me with this matter? The Integrand is: pi (t-6) P(t)= A sin _______ +B 12 The Integral is: a / | P(t)dt / b The following facts are known: A=0.56 B=0.28 a=4 b=20 Thank you guys! /Baz
 2011-04-10, 15:48 #2 Christenson     Dec 2010 Monticello 5·359 Posts I don't understand P(t), which I believe to be A * sin( X ) + B, where X is (t*pi-6)/12. Please confirm, also confirm that your interval of integration (a to b) is going backwards, that is, a > b, so dt is negative. Once we have that, the result will be straightforward.
 2011-04-10, 19:12 #3 Baztardo   Apr 2011 310 Posts Hi there! It´s pretty hard for me to explain. Perhaps I missplaced this question. I havent been to school for ages and I´m trying to solve this one in order to get som coordinates. I could explain the original problem: The formula above is edited by a person trying to solve it. Perhaps your more helped by the one originally made? pi (t-6) P(t)= A sin +B 12 b / | P (t) / a As before: A = 0,56 B = 0,28 a = 4 b = 20 I have this facts: N 57 43.Y00 E 012 54.X6Y By solving above integral Y and X should appear. The answer to the matter is a decimalnumber where Y is the numbers after the decimal "roughly". And X is half the number before the decimal. Englisg is myc second languagae, I hope you understand a little more than before :)
 2011-04-10, 19:56 #4 Mathew     Nov 2009 15E16 Posts $\int_{b}^aA\sin\left(\frac{\pi(t-6)}{12}\right)+B dt$ what I understood from Baz's post $\int_{a}^bA\sin\left(\frac{t\pi-6}{12}\right)+B dt$ what Christenson wrote Let us look at Christenson's equation without the points $\int_{}A\sin\left(\frac{t\pi-6}{12}\right)+Bdt$ Break this into two integrals $\int_{}A\sin\left(\frac{t\pi-6}{12}\right)dt+\int_{}Bdt$ Let us do the harder integral first $\int_{}A\sin\left(\frac{t\pi-6}{12}\right)dt$ A is a constant and can be placed outside of the integral $A\int_{}\sin\left(\frac{t\pi-6}{12}\right)dt$ Next is an u-substitution let $u=\frac{t\pi-6}{12}$ then $du=\frac{\pi}{12}dt$ Therefore $dt=\frac{12}{\pi}du$ Substitute the values $A\int_{}\frac{12}{\pi}\sin\left(u\right)du$ Again since $\frac{12}{\pi}$ is a constant move it out of the integral Now you have $A\frac{12}{\pi}\int_{}\sin\left(u\right)du$ Which is $-A\frac{12}{\pi}\cos\left(u\right)$ replace the u for what was substituted $-A\frac{12}{\pi}\cos\left(\frac{t\pi-6}{12}\right)$ Now do the second integral $\int_{}Bdt =Bt$ The whole equation is $-A\frac{12}{\pi}\cos\left(\frac{t\pi-6}{12}\right)+Bt$ Evaluating from a to b i.e. $\int_{a}^b$ it would be the following $\left(-A\frac{12}{\pi}\cos\left(\frac{b\pi-6}{12}\right)+B*b \right)-\left(-A\frac{12}{\pi}\cos\left(\frac{a\pi-6}{12}\right)+B*a\right)$ Entering in all the values =6.25624
 2011-04-10, 20:28 #5 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 9,433 Posts In 'Homework Help' forum, the usual thing to do would be to give hints, not a complete solution. One doesn't learn anything by simply copying the solution and turning it in.
 2011-04-10, 20:47 #6 Baztardo   Apr 2011 3 Posts Thanks! Thank you so much for helping me out with this one Mathew!! And as a respond to moderator. I wrote above that this might be places in wrong forum since i dont intend learning the steps from the answer, this is way way way beyond my current skills in mathematics. And second it wasn´t a homework task, rather some problemsolving task inside the geocaching community. Best wishes all of you! /Baz
 2011-04-10, 22:17 #7 Christenson     Dec 2010 Monticello 5×359 Posts In Baz' defense, he did state that he wasn't doing homework in the classical sense. Mr Stene, thanks for the nice formatting. Batalov, is my number theory homework, given to me by R.D Silverman (reading the literature and ensuring that I actually understand it, no grading involved) more appropriate here or under Factoring? I'm thinking about things like incremental QS, which may or may not work.

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