Question

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

Answer 1

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