Trig integral from hell

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.
 

lanedance

Homework Helper
3,305
2
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
 
31,930
3,893
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.
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).
 
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.
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top