$\int \sin x \sin 2x \sin 3x dx$ is equal to

The integral to evaluate is $\int \sin x \sin 2x \sin 3x dx$. This requires converting the product of sine functions into simpler terms.

---

Using the product-to-sum formula $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$ for $\sin x$ and $\sin 2x$:

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

$= \frac{1}{2}[\cos(-x) - \cos(3x)]$

$= \frac{1}{2}[\cos x - \cos 3x]$

---

The integral becomes:

$\sin x \sin 2x \sin 3x = \frac{1}{2}[\cos x - \cos 3x] \sin 3x$

$= \frac{1}{2}[\cos x \sin 3x - \cos 3x \sin 3x]$

---

Using $\cos A \sin B = \frac{1}{2}[\sin(A+B) - \sin(A-B)]$ for $\cos x \sin 3x$:

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

$= \frac{1}{2}[\sin 4x + \sin 2x]$

For $\cos 3x \sin 3x$:

$\cos 3x \sin 3x = \frac{1}{2}\sin(6x)$

---

Combining the terms:

$\sin x \sin 2x \sin 3x = \frac{1}{2}\left[\frac{1}{2}(\sin 4x + \sin 2x) - \frac{1}{2}\sin 6x\right]$

$= \frac{1}{4}\sin 2x + \frac{1}{4}\sin 4x - \frac{1}{4}\sin 6x$

---

Using $\int \sin(kx)dx = -\frac{\cos(kx)}{k} + C$:

$\int \sin x \sin 2x \sin 3x dx = \frac{1}{4}\int \sin 2x dx + \frac{1}{4}\int \sin 4x dx - \frac{1}{4}\int \sin 6x dx$

$= \frac{1}{4} \cdot \left(-\frac{\cos 2x}{2}\right) + \frac{1}{4} \cdot \left(-\frac{\cos 4x}{4}\right) - \frac{1}{4} \cdot \left(-\frac{\cos 6x}{6}\right) + C$

$= -\frac{1}{8}\cos 2x - \frac{1}{16}\cos 4x + \frac{1}{24}\cos 6x + C$

---

Converting to a common denominator of 48:

$= -\frac{6}{48}\cos 2x - \frac{3}{48}\cos 4x + \frac{2}{48}\cos 6x + C$

$= -\frac{1}{48}(6\cos 2x + 3\cos 4x - 2\cos 6x) + C$

Therefore, $\int \sin x \sin 2x \sin 3x dx = -\frac{1}{48}(6\cos 2x + 3\cos 4x - 2\cos 6x) + C$