#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#