What is #int sec(x)^3tan(x)dx# ?

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Answer 1 · 2018-08-07T22:35:32

#intsec(x)^3tan(x)dx=1/3sec(x)^3+C#, #C in RR#

#I=intsec(x)^3tan(x)dx#

#=-int(-sin(x))/(cos(x)^4)dx#

Let #u=cos(x)#

#du=-sin(x)dx#

So:

#I=-int1/u^4du#

#=1/(3u^3)#

#=1/(3cos(x)^3)+C#, #C in RR#

#=1/3sec(x)^3+C#, #C in RR#

\0/ Here's our answer !