How to integrate cos^4(3x+2)

Question

Hi, how do I integrate cos^4(3x+2)?

Answer 1

as cos(2a) = 2 cos^2(a) - 1, 2 cos^2(a) = cos(2a) + 1, or cos^2(a) = (cos(2a) + 1)/2

 

Thus, cos^4(3x+2) = (cos^2(3x+2))^2 =( (cos(6x+4) +1)/2)^2 = 1/4 cos^2(6x+4) + 1/2(cos(6x+4)) + 1/4

 

Then, 1/4 cos^2(6x+4) = 1/4(cos(12x + 8) + 1)/2 = 1/8 cos(12x+8) + 1/8

 

Substituting, we integrate 1/8 cos(12x+8) + 1/8 + 1/2(cos(6x+4)) + 1/4 =  1/8 cos(12x+8) + 1/2(cos(6x+4)) + 3/8

 

The integral is 1/96 sin(12x + 8) + 1/12 sin(6x+4) + 3/8 x + C