Integration problem.......?

Question

How do I intergrate 2cos^2 2x + 2cos^2 x . It is the 2x after the first cos^2 that I am unsure about.

Answer 1

The given expression equals 2 + cos4x + cos2x
( use cos2a =2cos^a -1). So the answer is 2x +sin4x/4+sin2x/2+ constant.

Answer 2

Ok, let's go on :

First let's remember :

cos2x = cos^2x - sin^2x

2cos^2(2x) = 1 + cos2x

cos4x = cos^2(2x) - sin^2(2x)

cos4x = 2cos^2(2x) - 1

1 + cos4x = 2cos^2(2x)

Integrate(1 + cos4x)dx + integrate(1 + cos2x)dx

easy then :

x + 1/4*sin(4x) + x + 1/2*sin(2x)

2x + 1/4*(sin4x + 2*sin(2x))

Hope that helps.

Answer 3

I = ∫ 2.cos² 2x dx+ ∫ 2.cos² x dx
Now cos 2x = cos ² x - sin ² x
cos 2x = 2 cos ²x - 1
2.cos² x = cos 2x + 1
2 cos² 2x = cos 4x + 1
I = ∫(cos 4x + 1).dx + ∫ (cos 2x + 1).dx
I = (1/4).sin 4x + x + (1/2).sin 2x + x + C
I = (1/4).sin 4x + (1/2).sin 2x + 2x + C