Antiderivate again..!?

Question

How to Antiderivate this ****..:)

2cos^2 (2x)

Thanks for helpin..

Answer 1

∫2cos^2 (2x) dx

Using the trig function cos(2x) = 2cos^2(x)-1
cos(4x) = 2cos^2(2x)-1
=> 2cos^2(2x) = 1+cos(4x)
= ∫1+cos(4x) dx - which can be done by inspection or substitution (u= 4x => du = 4dx)
= x + (1/4)sin(4x) + c

Answer 2

We know that cos^2(x) = (1 + cos(2x))/2, so that

Int cos^2(x) dx = Int (1 + cos(2x))/2 dx = 1/2 (x + 1/2 sin(2x) + C

Since the derivative of 2x is 2, it follows Int 2cos^2 (2x) dx = 1/2 (2x + 1/2 sin(4x)) + C = x + 1/4 sin(4x) + C where C is an integration constant

Answer 3

∫2cos^2 (2x) dx
= ∫1+cos(4x) dx
= x + (1/4)sin(4x) + c