Evaluate the integral (tan3x)dx

The solution to the problem is as follows:

tan (3x) = sin(3x) / cos(3x) 

let u = cos(3x), du = -3 sin(3x) 

Integrating -1/3 du / u = -1/3 ln (u) = -1/3 ln (abs (cos(3x)) + constant

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He is right i did this thing before

Use a calculator basically