Intuition Behind $ \frac{1}{\pi} \int_{0}^{\pi} (\dfrac{\sin(x + \sin^{-1}(\sin(x)*\alpha))}{\sin(x)})^{2} \,dx = 1$

Solution 1:

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:

$$\frac{1}{2 \pi}\int_0^{2 \pi} r^2 dx=\frac{1}{2 \pi}\int_0^{2 \pi} 1 dx = 1$$