The given function is \(\displaystyle \cos x \cdot \cos 2x \cdot \cos 3x \) and we need to compute the integral of this function with respect to \( x \), i.e., compute \(\displaystyle \int \cos x \cdot \cos 2 x \cdot \cos 3 x \, d x \).
Solution:
First, simplify the given function using trigonometric identities.
\( \begin{aligned} & \cos x \cdot \cos 2x \cdot \cos 3x \\ &= \frac{1}{2} [\cos(x + 2x) + \cos(x – 2x)] \cdot \cos 3x \quad \because \cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] \\ &= \frac{1}{2} [\cos 3x + \cos(-x)] \cdot \cos 3x \\ &= \frac{1}{2} [\cos 3x + \cos x] \cdot \cos 3x \\ &= \frac{1}{2} [\cos 3x + \cos x] \cdot \cos 3x \\ &= \frac{1}{2} \left[ \cos^2 3x + \cos x \cos 3x \right] \\ &= \frac{1}{2} \left[ \frac{1 + \cos 6x}{2} + \frac{1}{2} [\cos(x + 3x) + \cos(x – 3x)] \right] \quad \because \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \\ &= \frac{1}{2} \left[ \frac{1 + \cos 6x}{2} + \frac{1}{2} [\cos(x + 3x) + \cos(x – 3x)] \right] \\ &= \frac{1}{2} \left[ \frac{1 + \cos 6x}{2} + \frac{1}{2} [\cos 4x + \cos 2x]] \right] \\ &= \frac{1}{2} \left[ \frac{1 + \cos 6x}{2} + \frac{1}{2} (\cos 4x + \cos 2x) \right] \\ &= \frac{1}{2} \cdot \frac{1}{2} \left[ 1 + \cos 6x + \cos 4x + \cos 2x \right] \\ &= \frac{1}{4} \left[ 1 + \cos 2x + \cos 4x + \cos 6x \right] \\ \end{aligned} \)Now, integrate the given function:
\( \begin{aligned} &\int \cos x \cdot \cos 2x \cdot \cos 3x \, dx \\ &= \int \frac{1}{4} \left[ 1 + \cos 2x + \cos 4x + \cos 6x \right] dx \\ &= \frac{1}{4} \left[ \int 1 \, dx + \int \cos 2x \, dx + \int \cos 4x \, dx + \int \cos 6x \, dx \right] \\ &= \frac{1}{4} \left[ x + \frac{\sin 2x}{2} + \frac{\sin 4x}{4} + \frac{\sin 6x}{6} \right] + C, \quad \text{C is a constant} \\ \end{aligned} \)Thus, the integral of the given function is
\( \boxed{\int \cos x \cdot \cos 2x \cdot \cos 3x \, dx = \frac{1}{4} x + \frac{\sin 2x}{8} + \frac{\sin 4x}{16} + \frac{\sin 6x}{24} + C} \)Please let me know in the comments if you find any errors in this solution.