Question

Find the integrals of the function: \(cos\, 2x\, cos\, 4x\, cos\, 6x\)

Answer 1

Step 1: Use the identity \( \cos A \cos B = \frac{\cos(A+B) + \cos(A-B)}{2} \). First combine \( \cos 2x \) and \( \cos 4x \): \[ \cos 2x \cdot \cos 4x = \frac{\cos(6x) + \cos(-2x)}{2} = \frac{\cos 6x + \cos 2x}{2}. \]

Step 2: Multiply by \( \cos 6x \): \[ \cos 2x \, \cos 4x \, \cos 6x = \frac{1}{2} \left[ \cos 6x \cdot \cos 6x + \cos 2x \cdot \cos 6x \right]. \]

Step 3: Use \( \cos^2\theta = \frac{1+\cos 2\theta}{2} \) for the first term: \[ \cos 6x \cdot \cos 6x = \cos^2 6x = \frac{1 + \cos 12x}{2}. \] For the second term, again use the product-to-sum formula: \[ \cos 2x \cdot \cos 6x = \frac{\cos(8x) + \cos(-4x)}{2} = \frac{\cos 8x + \cos 4x}{2}. \]

Step 4: Combine results: \[ \cos 2x \cos 4x \cos 6x = \frac{1}{2} \left[ \frac{1 + \cos 12x}{2} + \frac{\cos 8x + \cos 4x}{2} \right]. \] Simplifying: \[ = \frac{1}{4} \left[ 1 + \cos 12x + \cos 8x + \cos 4x \right]. \]

Step 5: Integrate term by term: \[ \int \cos 2x \cos 4x \cos 6x \, dx = \frac{1}{4} \left[ \int 1 \, dx + \int \cos 12x \, dx + \int \cos 8x \, dx + \int \cos 4x \, dx \right]. \] \[ = \frac{1}{4} \left[ x + \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + \frac{\sin 4x}{4} \right] + C. \]

Final Answer: \[ \boxed{ \int \cos 2x \cos 4x \cos 6x \, dx = \frac{x}{4} + \frac{\sin 12x}{48} + \frac{\sin 8x}{32} + \frac{\sin 4x}{16} + C } \]

Answer 2

The correct answer is: \(=\frac{1}{4}[\frac{sin12x}{12}+\frac{sin8x}{8}+x+\frac{sin4x}{4}]+C\)
It is known that, \(cosA\,cosB = \frac{1}{2}[cos(A+B)+cos(A-B)]\)
\(∴ ∫cos2x(cos4x\,cos6x)dx = ∫cos2x[\frac{1}{2}{cos(4x+6x)+cos(4x-6x)}]dx\)
\(=\frac{1}{2} ∫ {cos2x\,cos10x+cos2x\,cos(-2x)}dx\)
\(=\frac{1}{2} ∫[cos2x\,cos10x+cos^2 2x]dx\)
\(=\frac{1}{2} ∫[{\frac{1}{2}cos(2x+10x)+cos(2x-10x)}+(\frac{1+cos4x}{2})]dx\)
\(=\frac{1}{4} ∫(cos12x+cos8x+1+cos4x)dx\)
\(=\frac{1}{4}[\frac{sin12x}{12}+\frac{sin8x}{8}+x+\frac{sin4x}{4}]+C\)