Integration help!!?

Question

integral cos^4(2x) dx

Answer 1

Integral (cos^4(2x) dx )

Note that this is the same as [cos^2(2x)]^2.

Integral ( [cos^2(2x)]^2 dx )

Now, use the half angle identity; cos^2(y) = (1/2)(1 + cos2y), so

Integral { ( [(1/2) (1 + cos[2(2x)]) )^2 dx }

Simplifying, we have

Integral { (1/4) (1 + cos(4x))^2 dx }

Pulling the (1/4) out of the integral,

(1/4) * Integral ( [1 + cos(4x)]^2 dx )

Squaring the binomial,

(1/4) * Integral ( [1 + 2cos(4x) + cos^2(4x)] dx )

We apply the half angle identity once more, this time to cos^2(4x).

(1/4) * Integral ( [1 + 2cos(4x) + (1/2)(1 + cos(8x)) ] dx )

(1/4) * Integral ( [1 + 2cos(4x) + (1/2) + (1/2)cos(8x) ] dx )

Now, it's simply a matter of integrating. The integral of something in the form cos(nx) will be (1/n)sin(nx). This leaves us with

(1/4) * [x + 2(1/4)sin(4x) + (1/2)x + (1/2)(1/8)sin(8x)] + C

Distributing the 1/4, we have

(1/4)x + (2/16)sin(4x) + (1/2)x + (1/16)sin(8x) + C

Notice that we can combine (1/4)x + (1/2)x into (3/4)x.

(3/4)x + (1/8)sin(4x) + (1/16)sin(8x) + C

Answer 2

Integrate cos^4(2x) with respect to x.

First rearrange the terms to make it easier to integrate.

cos^4(2x) = {[1 + cos(4x)]/2}² = (1/4){1 + 2cos(4x) + cos²(4x)}
= 1/4 + cos(4x)/2 + (1/4)[1 + cos(8x)]/2
= 1/4 + cos(4x)/2 + 1/8 + cos(8x)/8
= 3/8 + cos(4x)/2 + cos(8x)/8

Now we can integrate.

∫cos^4(2x) dx = ∫{3/8 + cos(4x)/2 + cos(8x)/8}dx
= 3x/8 + sin(4x)/8 + sin(8x)/64 + C