What is the antiderivative of #sec^5xtan^7x#?

1 Answer
Mar 22, 2015

The answer is: #sec^11x/11-sec^9x/3+(3sec^7x)/7-sec^5x/5+c#.

Remembering that:

  • #sec^2x=1/cos^2x=(sin^2x+cos^2x)/cos^2x=#

#=sin^2x/cos^2x+cos^2x/cos^2x=tan^2x+1rArr#

#tan^2x=sec^2x-1#;

  • #intsecxtanxdx=secx+c#;

  • #int[f(x)]^n*f'(x)dx=[f(x)]^(n+1)/(n+1)+c#.

Than:

#intsec^5x*tan^7xdx=intsec^5x*tan^6x*tanxdx=#

#=intsec^5x*(sec^2x-1)^3*tanxdx=#

#=intsec^5x(sec^6x-3sec^4x+3sec^2x-1)*tanxdx=#

#=int(sec^11xtanx-3sec^9xtanx+3sec^7xtanx-sec^5xtanx)dx=#

#=intsec^10xsecxtanxdx-3intsec^8xsecxtanxdx+#

#+3intsec^6xsecxtanxdx-intsec^4xsecxtanxdx=#

#=sec^11x/11-(3sec^9x)/9+(3sec^7x)/7-sec^5x/5+c=#

#=sec^11x/11-sec^9x/3+(3sec^7x)/7-sec^5x/5+c#.