Question

cos2x.cos4x.cos6x,Integrate the given function defined on proper domain w.r.t. x.

Answer 1

so, COS2θ × COS4θ = 1/2 (cos6θ + cos2θ)

Answer 2

question is ---> $\int{cos2x.cos4x.cos6x}\,dx$

we know, 2cosC.cosD = cos(C + D) + cos(C - D)
so, cos2x.cos4x = 1/2[2cos2x.cos4x]
= 1/2[cos(2x + 4x) + cos(4x - 2x)]
= 1/2[cos6x + cos2x]

now, cos2x.cos4x.cos6x = 1/2[cos6x + cos2x]cos6x
= 1/2[cos6x.cos6x + cos6x.cos2x]
= 1/2[1/2(cos12x + cos0°) + 1/2(cos8x + cos4x)]
= 1/4[1 + cos12x + cos8x + cos4x]

now,
$\int{\frac{1}{4}[1+cos12x+cos8x+cos4x]}\,dx\\\\\\=\int{dx}+\int{cos12x}\,dx+\int{cos8x}\,dx+\int{cos4x}\,dx\\\\\\=x+\left[\frac{sin12x}{12}\right]+\left[\frac{sin8x}{8}\right]+\left[\frac{sin4x}{4}\right]+C$