4cos^3(2x) dx Evaluate the integral?

Answer 1

factor out a cos^2(2x)
Then use the trig identity sin^2(x) + cos^2(x) = 1
or in this case it's 2x instead of x inside the sin and cos.
It should be pretty easy from there.

Answer 2

Use the trigonometry formula
cos (3x) = 4cos^3(x) - 3 cos(x)
or 4cos^3(x) = cos (3x) + 3 cos(x)
or cos^3(x) =1/4 cos (3x) + 3/4 cos(x)
cos^3(2x) =1/4 cos (6x) + 3/4 cos(2x)
Taking integration
= 1/4 {-sin (6x)/6 + 3/4{ -sin (2x)}/2 +C
= -(1/24) sin (6x) - 3/8 { sin (2x)} + C

Answer 3

Here is the solution:
in integral 4cos^3(2x), we substitute 2x=u, so, du=2dx. we write du/2=dx. we rewrite the original expression in terms of u: 4cos^3(u)du/2=2cos^3(u)du. god, i'm tired of writing the whole solution, here is the answer: 2sin(2x)-2/3sin^3(2x). hope I have given you enough hints to solve this problem.

Answer 4

(1/6)(9sin(2x)+sin(6x))