cuttlefish said:
Homework Statement
\int \frac{3cosx-10sinx}{10cosx+3sinx} from -pi/3 to pi/3
Homework Equations
some sort of identity maybe?
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.
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).
