#intcos^4xdx#?

Question

383 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-19T11:01:28

# 1/32(12x+8sin2x+sin4x)+C#.

We have, #cos^4x=(cos^2x)^2={1/2(1+cos2x)}^2#,

#=1/4(1+2cos2x+cos^2 2x)#,

#=1/4{1+2cos2x+1/2(1+cos4x)}#,

#=1/8(3+4cos2x+cos4x)#.

#:. intcos^4xdx=1/8int(3+4cos2x+cos4x)dx#,

#=1/8(3x+4*1/2sin2x+1/4sin4x)#.

#rArr intcos^4xdx=1/32(12x+8sin2x+sin4x)+C#.