∫dx/(cos^2(3-2x))=

1)2tg(3-2x) + C

2)2ctg(3-2x) + C

3)(1/2)ctg(3-2x) + C

4)-(1/2)tg(3-2x) + C

To find the antiderivative of the given function ∫dx/(cos^2(3-2x)), we can use the trigonometric identity: 1 + tan^2(x) = sec^2(x).

Let's rewrite the integral using this identity:
∫dx/(cos^2(3-2x)) = ∫sec^2(3-2x)dx

Now, we can use the chain rule to integrate this function. Let u = 3-2x, then du = -2dx.

Substitute u = 3-2x and du = -2dx into the integral:
∫sec^2(u)(-du/2) = (-1/2)∫sec^2(u)du

Integrating sec^2(u)du gives us tan(u), so the integral becomes:
(-1/2)tan(u) + C

Substitute back u = 3-2x:
(-1/2)tan(3-2x) + C

To simplify the result, recall that tan(x) = sin(x)/cos(x):
(-1/2)(sin(3-2x)/cos(3-2x)) + C
= -(1/2)(sin(3-2x)/cos(3-2x)) + C

Therefore, the antiderivative of ∫dx/(cos^2(3-2x)) is:
-(1/2)(sin(3-2x)/cos(3-2x)) + C