What is the integral of #int tan(2x) dx#?

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Answer 1 · 2016-02-19T15:23:47

#inttan(2x)dx =-1/2ln|cos(2x)|+C#

We will proceed using substitution, along with the fact that #int1/xdx = ln|x|+C#

Let #u = cos(2x) => du = -2sin(2x)dx#

Then we have

#inttan(2x)dx = intsin(2x)/cos(2x)dx#

#=-1/2int1/cos(2x)(-2sin(2x))dx#

#=-1/2int1/udu#

#=-1/2ln|u|+C#

#=-1/2ln|cos(2x)|+C#