# Integral of cosine function

• daudaudaudau

#### daudaudaudau

Hi. I have been experimenting a little to come up with the following "conjecture"
$$\int_0^{2\pi}d\phi f(a+b\cos\phi)\sin\phi=0$$
where a and b are arbitrary constants and f(x) is any function. Is this true? I guess it can be shown by expanding f in a power series of cosines?

hi daudaudaudau!

doesn't work for f = √, a = b

Yes, I'm getting that that this is true. It can be proven using the substitution
$$x=\cos{\phi}$$
$$dx=-\sin{\phi}d\phi$$
But this substitution is not 1-to-1: Each x value corresponds to 2 phi values on [0,2pi]. So you need to break the region of integration into [0,pi] and [pi,2pi]. If you look at the graph of the cos function, you will see it is 1-to-1 on these two intervals and goes from 1 to -1 on [0,pi] and from -1 to 1 on [pi,2pi]. So the integral becomes:
$$\int_0^{2\pi}d\phi f(a+b\cos\phi)\sin\phi= \int_1^{-1}-f(a+bx)dx + \int_{-1}^{1}-f(a+bx)dx = \int_{-1}^{1}f(a+bx)dx + \int_{-1}^{1}-f(a+bx)dx = 0$$

Tiny-tim: I get this even in the example you gave.

hmm … i think I've been misled by the ambiguity of the √ function

yes, we can get it directly from the original integral …

if f = g', then it's ∫ [g(a + bcosφ)]' dφ, = [g(a + bcosφ)]φ=0

hmm … i think I've been misled by the ambiguity of the √ function

yes, we can get it directly from the original integral …

if f = g', then it's ∫ [g(a + bcosφ)]' dφ, = [g(a + bcosφ)]φ=0

Happens to everyone
That is a better way of proving it.

hmm … i think I've been misled by the ambiguity of the √ function

yes, we can get it directly from the original integral …

if f = g', then it's ∫ [g(a + bcosφ)]' dφ, = [g(a + bcosφ)]φ=0

What happened to the sine function?

chain rule

clever :-)

The integral from -pi to pi is zero, because you are integrating an odd function.

And the integral from -pi to pi equals the integral from 0 to 2pi.

QED.