How do I find the antiderivative of #sec^3 x tan^4 x dx#?

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Answer 1 · 2015-03-23T18:38:19

I'm going to guess that you'll need trigonometric power reduction formulas and integration by parts and perhaps integration by substitution.
I'm guessing this based on an answer by of:

#1/96[1/8(78sinx-47sin3x+3sin5x)sec^6x-6ln(cos(x/2)-sin(x/2))+6ln(cos(x/2)+sin(x/2))]#.

from WolframAlpha.

WolframAlpha is not particularly efficient though.

For example: #-6ln(cos(x/2)-sin(x/2))+6ln(cos(x/2)+sin(x/2))=-6ln((1-sinx)/cosx)#

So the answer could be written:

#1/96[1/8(78sinx-47sin3x+3sin5x)sec^6x-6ln((1-sinx)/cosx)]#.

Good luck.