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

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Answer 1 · 2016-08-30T19:30:13

#1/2tan(2x)-x+C#

Note that #(tan(2x))^2=tan^2(2x)#.

We have:

#inttan^2(2x)dx#

Recall that, through the Pythagorean identity, #tan^2(x)=sec^2(x)-1#.

#=int(sec^2(2x)-1)dx#

Split up the integral:

#=intsec^2(2x)dx-intdx#

#=intsec^2(2x)dx-x#

Now, let #u=tan(2x)#. This means that #du=2sec^2(2x)dx#.

#=1/2int2sec^2(2x)dx-x#

#=1/2intdu-x#

#=1/2u-x+C#

#=1/2tan(2x)-x+C#