Having problems evaluating definite integral

[tjkubo]

Homework Statement

I'm trying to evaluate
$$\int_0^{2\pi} \frac{9-6\cos t}{5-4\cos t} dt \ .$$
I found that the antiderivative of the integrand is
$$F(x) = \frac{3}{2}x + \tan^{-1} \left(3\tan\frac{x}{2} \right)$$
so using the FTC, the integral should be
$$F(2\pi)-F(0)=3\pi$$
But using a calculator to evaluate the integral yields $4\pi$.
What am I doing wrong here?

Homework Equations

The Attempt at a Solution

 

[Sourabh N]

Hint: tan^-1 0 is not always equal to 0.

 

[tjkubo]

I don't understand what you mean. If tan-1 x were multivalued, then it would not be a function, so I implicitly restricted tan-1 x to its principal branch.

 

[Ray Vickson]

Homework Statement

I'm trying to evaluate
$$\int_0^{2\pi} \frac{9-6\cos t}{5-4\cos t} dt \ .$$
I found that the antiderivative of the integrand is
$$F(x) = \frac{3}{2}x + \tan^{-1} \left(3\tan\frac{x}{2} \right)$$
so using the FTC, the integral should be
$$F(2\pi)-F(0)=3\pi$$
But using a calculator to evaluate the integral yields $4\pi$.
What am I doing wrong here?

Homework Equations

The Attempt at a Solution

You need to worry about whether or not the hypotheses of the fundamental theorem of calculus hold; that is, whether or not $\int_0^{2 \pi} f(x) \, dx = F(2\pi) - F(0)$ actually holds (where F is an antiderivative of f). If F(x) sweeps over points arising from different branches of 'arctan' as x goes from 0 to 2π, you might run into trouble. To be safe, break up the integral as $I = I_1 + I_2,$ where x goes from 0 to π in the first integral and from π to 2π in the second; then change variables to x = π + y in the second integral, so you again have an integral from 0 to π.

RGV

 

[tjkubo]

Aren't the hypotheses of the FTC just that F is differentiable and F'=f? These seem to hold for the function I gave, so I don't see exactly where the problem is.

 

[SammyS]

Aren't the hypotheses of the FTC just that F is differentiable and F'=f? These seem to hold for the function I gave, so I don't see exactly where the problem is.

Yes, you need F to be differentiable.

If $\displaystyle F(x) = \frac{3}{2}x + \tan^{-1} \left(3\tan\frac{x}{2} \right)\,,$ notice that F has jump discontinuities at x = (2k+1)π. In other words, F(x) is not differentiable when x is an odd integer multiple of π.

This requires you to break up the integral as has been suggested.