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Trig integral from hell

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int[/tex] [tex]\frac{3cosx-10sinx}{10cosx+3sinx}[/tex] from -pi/3 to pi/3


    2. Relevant equations
    some sort of identity maybe?


    3. The attempt at a solution
    well...I made the mistake of saying the answer was zero since it's an odd function (yes, i know this is ridiculously not right) and so incurred the wrath of my calculus teacher. if someone can help me get started on this then maybe I can salvage something resembling a grade.
     
  2. jcsd
  3. Mar 9, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    you could try writing
    [tex] 10 = A.cos(\alpha) [/tex]
    [tex] 3 = A.sin(\alpha) [/tex]

    for unknown A & alpha, plug this into your equation then make use of some tirg identities for angle sums...
    maybe not that exactly... but something along those lines has a good chance of working
     
  4. Mar 9, 2009 #3

    Mark44

    Staff: Mentor

    lanedance had the start of this idea. Let's see if I can finish it off. First off, notice that the same two numbers appear: 3 and 10. If you square them, add them, and take the square root, you get sqrt(109). This will enter into things in a little while.

    1. For the numerator, you have 3cos(x) - 10 sin(x). If this were sinA*cosx - cosA * sinx, you could rewrite this as sin(A - x).

    2. Simililarly, you have 10cos(x) + 3sin(x) in the denominator. If this were cosA*cos(x) + sinA*sin(x), you could rewrite it as cos(A - x).

    The trouble is, there is no A for which cosA = 10 or for which sinA = 3. This is where sqrt(109) comes in. Multiply the numerator and denominator of your integrand by 1 in the form of (1/sqrt(109))/(1/sqrt(109)).

    The numerator now looks like 3/sqrt(109) * cos(x) - 10/sqrt(109)* sin(x). The denominator now looks like 10/sqrt(109) * cos(x) + 3/sqrt(109) * sin(x). So sinA = 3/sqrt(109) and cosA = 10/sqrt(109), and these should hold for both the numerator and denominator.

    With this work, you can rewrite the integrand as sin(A - x)/cos(A - x), which is tan(A - x), which is easy enough to integrate. You'll need to find A, of course, which can be done by using the appropriate inverse functions in either sinA = 3/sqrt(109) or cosA = 10/sqrt(109).
     
  5. Mar 9, 2009 #4
    wow, thanks you guys. this is definitely something that I never would have done. we spent like two seconds on product sum rules and I didn't really understand that they could be used like this. i'll need to write this down and stare at it for a couple minutes but thanks! I sense a passing grade in my future.
     
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