The integral of tan (2x) has to be determined.

Int [tan (2x) dx]

let 2x = y

dy = 2 dx

=> Int[ (tan y)(1/2) dy]

=> (1/2)*Int[ tan y dy]

=> (1/2)*Int[ (sin y)/(cos y) dy]

let cos y= t

dt = -sin y dy

=> (1/2)*Int [ (-1/ t) dt]

=> (1/2)*-ln t + C

=> [ln(1/t)]/2 + C

substitute t = cos y

=> [ln(1/ (cos y))]/2 + C

=> [ln (sec y)] / 2 + C

substitute y = 2x

=> [ln (sec (2x))] / 2 + C

**The required integral is [ln (sec 2x)] / 2 + C**

First, we'll substitute 2x by t.

2x = t

Differentiating both sides, we'll get:

2dx = dt => dx = dt/2

We'll re-write the integral:

Int tan(2x)dx = Int (tan t)*dt/2

We'll write tan t = sin t/cos t

Int (tan t)*dt/2 = Int (sin t/cos t)*(dt/2)

We'll apply substitution technique to solve the indefinite integral. We'll put cos t = v.

- sin t dt = dv

Int (sin t/cos t)*(dt/2) = Int -dv/2v = - (1/2)*ln |v| + C

Int (tan t)*dt/2 = - (1/2)*ln |cos t| + C

Int tan (2x)*dx = - (1/2)*ln |cos (2x)| + C

Int tan (2x)*dx = ln [1/sqrt (cos 2x)] + C

**The indefinite integral of the given function is: Int tan (2x)*dx = ln [1/sqrt (cos 2x)] + C**