# Confirm definite integral equals zero $\frac{\sin(x)}{(1-a\cos(x))^{2}}$

Is this statement about the definite integral of a particular function $F$ true? $$\int_0^{2\pi}F(x)\, \mathrm{d}x = \int_0^{2\pi}\frac{\sin(x)}{(1-a\cos(x))^2}\, \mathrm{d}x = 0 \ \text{ for }\ 0<a<1$$

I have evaluated this expression (in WolframAlpha) for various values of a and they all give the value zero. I have read that integrals of the form $$\int G(\cos(x))\sin(x)\, \mathrm{d}x$$ where $G$ is some continuously integrable function are zero over the range $-\pi/2$ to $\pi/2$.

(Edited after comment from Andrey) It seems possible to proceed from here to confirm the postulated statement by symmetry. The function $F$ to be integrated is cyclic with period 2$\pi$ such that $F(x-2\pi) = F(x) =F(x+2\pi)$. Then we just need to prove that the two integrals:- (1) between $-\pi$ and $-\pi/2$, and (2) between $\pi/2$ and $\pi$ are equal in magnitude and opposite in sign.

This would be the case if $F(x) = -F(-x)$. Such is actually the case because the denominator in F() expands to $(1-2a\cos(x)+a^2\cos^2x)$ and has the same values for $(+x)$ and $(-x)$. Whereas the numerator $\sin(x)$ is such that $\sin(x) = -\sin(-x)$.

However I would still like to find an analytical solution.

• It is the case that $F(x)=-F(x)$. The denominator $1-2a\cos x + a^2\cos^2x$ for $x<0$ is the same as $x>0$, since cosine is an even function, i.e. $\cos(-x) = \cos(x)$. – Andrey Kaipov Oct 11 '14 at 22:09
• @Andrey Yes you are right of course. Doh! – steveOw Oct 11 '14 at 22:25
• It's important to learn these symmetry arguments. There is a famous integral from the Putnam that cannot be done any other way: $\int_0^{\pi/2} \frac{1}{1+\tan^{\sqrt{2}} x} = \frac{\pi}{4}$. – Slade Oct 12 '14 at 0:59

Let $u=1-a\cos x$, $du=a\sin x dx$ to get $\displaystyle\frac{1}{a}\int_{b}^{b}\frac{1}{u^2}du=0$ $\;\;\;$ (where $b=1-a$).

• I dont understand the limits b..b. – steveOw Oct 11 '14 at 22:42
• When $x=0, u=1-a(1)=1-a$ and when $x=2\pi, u=1-a(1)=1-a$. – user84413 Oct 11 '14 at 22:51
• @user844413 Wow that is so slick! – steveOw Oct 11 '14 at 23:00

Using the general rule (link)

$$\int_a^bF(x)dx = \int_a^bF(a+b-x)dx,$$

we have in this case

$$\int_0^{2\pi}F(x)dx = \int_0^{2\pi}F(2\pi-x)dx,$$

and, from knowledge of the symmetries of the sin and cos functions, we know in this case that $$F(x) = - F(2\pi-x),$$

so, with $I =\int_0^{2\pi}F(x)dx$, we have

$$I=-I,$$

which can only be true if $$I = 0$$

• But how do I know your two integrals are equivalent to start with? I only know: F(x)=-F(-x) and F(x)=-F(2pi-x) and hence F(-x)=F(2pi-x). – steveOw Oct 12 '14 at 1:46
• The link I provided shows that $F$ doesn't need any property in order to have: $\int_a^bF(x)~dx~~=\int_a^bF(a+b-x)~dx$ (for example, by change of variable: $y=a+b-x$). The lucky part for your integral is that we get $-I$ for the second integral. – ir7 Oct 12 '14 at 1:57
• (Aha, I didn't spot the link). The very useful equation in your comment is fundamental to this answer. I suggest it is included in the answer. – steveOw Oct 12 '14 at 15:03
• Ok.It looks better now. Cheers. :) – ir7 Oct 12 '14 at 15:25

This is a special case of a general fact about $u$-substitution. If $G(x)$ is integrable on the interval $[a,b]$, with antiderivative $g(x)$, and $u$ is differentiable, then $$\int_a^b G(u(x))\,u'(x)\, dx = g(u(b))-g(u(a)).$$

If $u(a)=u(b)$, the integral is zero.

The integrand in your example has this form, where $u(x)=\cos(x)$ and $G(u)=\frac{-1}{(1-a\cos(x))^2}$, and $\cos(x)$ has the same value at both limits of integration, so the integral is zero. (You can apply the substitution user84413 suggested, or the simpler $u=\cos x$ to show it.)

You can write down lots of messy-looking integrals that turn out to be zero because they have this form for some $u(x)$ and $G(x)$.

$$\int_1^3 e^{x^2-4x+7}(2-x)\, dx$$

$$\int_0^{2\pi} (\pi-x)\log(2+\sin^2(x-\pi)^2)\, dx$$

$$\int_{\pi/2}^{3\pi/2} (\cos^2 x)^{\sin^2 x}\sin2x\,dx$$

• I like @ir7’s trick for this, which works for these examples. Here’s one where you can’t quickly show that $I=-I$ that way: $\displaystyle\int_1^9 (4\sqrt x-x)^{4\sqrt x-x}(\frac{2}{\sqrt{x}}-1)\,dx$. – Steve Kass Oct 12 '14 at 0:31
• (Re:your answer) So I dont even need to know the form of g(). Nice. – steveOw Oct 12 '14 at 1:01