We are asked to evaluate the definite integral $\int_{0}^{\frac{\pi}{6}} \frac{\sin x \sin(x + \frac{\pi}{3}) \sin(x + \frac{2\pi}{3})}{\sin 3x + \cos 3x} dx$.

1. Problem Description

We are asked to evaluate the definite integral

$\int_{0}^{\frac{\pi}{6}} \frac{\sin x \sin(x + \frac{\pi}{3}) \sin(x + \frac{2\pi}{3})}{\sin 3x + \cos 3x} dx$.

2. Solution Steps

We first simplify the numerator.

Using the product-to-sum identity,

$\sin a \sin b = \frac{1}{2}[\cos(a-b) - \cos(a+b)]$, we have

$\sin(x + \frac{\pi}{3}) \sin(x + \frac{2\pi}{3}) = \frac{1}{2}[\cos(-\frac{\pi}{3}) - \cos(2x + \pi)] = \frac{1}{2}[\cos(\frac{\pi}{3}) - \cos(2x + \pi)] = \frac{1}{2}[\frac{1}{2} - (-\cos 2x)] = \frac{1}{4} + \frac{1}{2} \cos 2x$.

Then the numerator becomes

$\sin x (\frac{1}{4} + \frac{1}{2} \cos 2x) = \frac{1}{4}\sin x + \frac{1}{2} \sin x \cos 2x = \frac{1}{4} \sin x + \frac{1}{2} \sin x (1-2\sin^2 x) = \frac{1}{4}\sin x + \frac{1}{2} \sin x - \sin^3 x = \frac{3}{4} \sin x - \sin^3 x$.

We can use the triple angle formula $\sin 3x = 3\sin x - 4\sin^3 x$, which implies $\sin^3 x = \frac{3\sin x - \sin 3x}{4}$.

Therefore, the numerator is $\frac{3}{4} \sin x - \frac{3\sin x - \sin 3x}{4} = \frac{3\sin x - 3\sin x + \sin 3x}{4} = \frac{1}{4}\sin 3x$.

The integral becomes $\int_{0}^{\frac{\pi}{6}} \frac{\frac{1}{4} \sin 3x}{\sin 3x + \cos 3x} dx$.

Let $u = 3x$. Then $du = 3dx$, so $dx = \frac{1}{3} du$.

When $x=0$, $u=0$. When $x=\frac{\pi}{6}$, $u = 3(\frac{\pi}{6}) = \frac{\pi}{2}$.

The integral becomes $\int_{0}^{\frac{\pi}{2}} \frac{\frac{1}{4}\sin u}{\sin u + \cos u} \frac{1}{3} du = \frac{1}{12} \int_{0}^{\frac{\pi}{2}} \frac{\sin u}{\sin u + \cos u} du$.

Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin u}{\sin u + \cos u} du$.

Using the substitution $u = \frac{\pi}{2} - v$, $du = -dv$.

When $u=0$, $v=\frac{\pi}{2}$. When $u=\frac{\pi}{2}$, $v=0$.

$I = \int_{\frac{\pi}{2}}^{0} \frac{\sin(\frac{\pi}{2} - v)}{\sin(\frac{\pi}{2} - v) + \cos(\frac{\pi}{2} - v)} (-dv) = \int_{0}^{\frac{\pi}{2}} \frac{\cos v}{\cos v + \sin v} dv$.

So $I = \int_{0}^{\frac{\pi}{2}} \frac{\cos u}{\cos u + \sin u} du$.

Therefore, $2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin u}{\sin u + \cos u} du + \int_{0}^{\frac{\pi}{2}} \frac{\cos u}{\sin u + \cos u} du = \int_{0}^{\frac{\pi}{2}} \frac{\sin u + \cos u}{\sin u + \cos u} du = \int_{0}^{\frac{\pi}{2}} 1 du = [u]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}$.

Thus, $I = \frac{\pi}{4}$.

The original integral becomes $\frac{1}{12} \cdot \frac{\pi}{4} = \frac{\pi}{48}$.

3. Final Answer

$\frac{\pi}{48}$