How do you find the integral of #int tan^3(2x) sec^5(2x) dx#?

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Answer 1 · 2015-10-17T14:05:11

#I = 1/2 [(sec^7 2x)/7-(sec^5 2x)/5] + C#

#sec2x=t => sec2xtan2x*(2x)'dx=dt#

#2sec2xtan2xdx=dt#

#sec^2 2x=1+tan^2 2x => tan^2 2x = sec^2 2x -1#

#I = int tan^3 2x sec^5 2x dx#

#I = int tan^2 2x sec^4 2x tan2x sec 2x dx#

#I = 1/2 int (sec^2 2x -1) sec^4 2x (2tan2x sec 2x dx)#

#I = 1/2 int (t^2 -1) t^4 dt = 1/2 int (t^6 - t^4) dt = 1/2 [t^7/7-t^5/5] + C#

#I = 1/2 [(sec^7 2x)/7-(sec^5 2x)/5] + C#