How do you integrate #sec2x tan2x dx#?

1 Answer
Gió
Feb 21, 2015

I would write the integral as:

#intsec(2x)tan(2x)dx=int1/cos(2x)sin(2x)/cos(2x)dx=#

we can see that:

#d[cos(2x)]=-2sin(2x)dx#
and so:
#sin(2x)dx=(d[cos(2x)])/-2#

with this in mind I write the integral as:

#int1/cos^2(2x)*1/-2d[cos(2x)]=#
#=1/-2intcos^(-2)(2x)d[cos(2x)]=# (treating #cos(2x)# as if it were #x# in a normal integration where you get #x^(n+1)/(n+1)#)
#=1/2*1/cos(2x)+c#

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