What is #inttan(x)^3sec(x)^3dx# ?

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Answer 1 · 2018-08-07T22:19:38

#inttan(x)^3sec(x)^3dx=1/5sec(x)^5-1/3sec(x)^3+C#

#I=inttan(x)^3sec(x)^3dx#

#=inttan(x)tan(x)^2sec(x)^3dx#

Because #tan(x)^2=sec(x)^2-1#

#I=inttan(x)sec(x)^3(sec(x)^2-1)dx#

#=inttan(x)sec(x)^5dx-inttan(x)sec(x)^3dx#

#=-int(-sin(x))/(cos(x)^6)dx+int(-sin(x))/(cos(x)^4)dx#

Let #u=cos(x)#
#du=-sin(x)dx#

So:

#I=int1/u^4du-int1/u^6du#

#=1/(5u^5)-1/(3u^3)+C#, #C in RR#

#=1/(5cos(x)^5)-1/(3cos(x)^3)+C#, #C in RR#

#=1/5sec(x)^5-1/3sec(x)^3+C#, #C in RR#