Intuition Behind $ \frac{1}{\pi} \int_{0}^{\pi} (\dfrac{\sin(x + \sin^{-1}(\sin(x)*\alpha))}{\sin(x)})^{2} \,dx = 1$ Given circle with R = $1$ and $B$ the center. let's define a point $A$ which $AB=a$ (between $0$ and $1$). And let have a point $C$ on the circumference, and let $\angle BAC=x$.

We can easily calculate the side $AC$ from the triangle $ABC$ since we have two sides and an angle (though not between them); according to my calculation it results in $\angle C = \sin^{-1}(a * \sin(x))$  and from there it follows that $AC = \dfrac{\sin(x + \sin^{-1}(\sin(x)*a))}{\sin(x)}$. If we have then the point $C$ running all over the circle with uniform angular speed with respect to $A$ (not $B$), and we want the average $AC^2$, we find it to be $1$ no matter what $a$ is (in this link I took $a=0.27$ but any $a$ between $0$ and $1$ could be taken]
$$ \frac{1}{\pi} \int_{0}^{\pi} (\dfrac{\sin(x + \sin^{-1}(\sin(x)*a))}{\sin(x)})^{2} \,dx = 1$$
I find this result not so trivial. So I wonder is there any kind of intuition or reasoning (I mean without calculating the integral) that we could tell in advanced that the average $AC^2$ will end up to be exactly $1$?
 A: I don't see a geometrical reason, but one can explain it by a simpler integral calculation, due to different cancellations as we shall see.
Let us start by applying cosine rule to angle $A$ in triangle $ABC$. Assuming that the circle has radius $1$:
$$1=r^2 + a^2-2 a r \cos x \ \iff \ 1=(r-a \cos x)^2\underbrace{-a^2 (\cos x)^2 + a^2}_{+a^2 (\sin x)^2}$$
$$(r-a \cos x)^2 = 1 - a^2 (\sin x)^2$$
giving
$$r=a \cos x + \sqrt{1-a^2 (\sin x)^2}\tag{1}$$
which is the polar equation of a shifted circle.
As a consequence, squaring (1):
$$r^2=1 \ \underbrace{-a^2 (\sin x)^2+a^2(\cos x)^2}_{T_1=a^2 \cos(2x)}+\underbrace{2 a \cos x \sqrt{1-a^2 (\sin x)^2}}_{T_2} \tag{2}$$
In fact terms $T_1$ and $T_2$ will disappear in an integration of (2) from $0$ to $2\pi$. Indeed,

*

*Term $T_1$  contributes to... $0$ (because we integrate periodical function cosine on 2 periods): $\int_0^{2 \pi}\cos(2x) dx = 0$.


*Term $T_2$ has also a $0$ global contribution because, if we divide the integration interval into two parts, we see that
$$\int_{\pi}^{2 \pi}T_2 dx \ \ \text{is the opposite of} \ \  \int_{0}^{\pi}T_2 dx$$
(due to relationships $\cos(x+\pi)=-\cos(x), \sin(x+\pi)=-\sin(x)$).
Therefore, (2) gives the following expected mean value:
$$\frac{1}{2 \pi}\int_0^{2 \pi} r^2 dx=\frac{1}{2 \pi}\int_0^{2 \pi} 1 dx = 1$$
