What is the antiderivative of sec^5xtan^7xsec5xtan7x?

1 Answer
Mar 22, 2015

The answer is: sec^11x/11-sec^9x/3+(3sec^7x)/7-sec^5x/5+csec11x11sec9x3+3sec7x7sec5x5+c.

Remembering that:

  • sec^2x=1/cos^2x=(sin^2x+cos^2x)/cos^2x=sec2x=1cos2x=sin2x+cos2xcos2x=

=sin^2x/cos^2x+cos^2x/cos^2x=tan^2x+1rArr=sin2xcos2x+cos2xcos2x=tan2x+1

tan^2x=sec^2x-1tan2x=sec2x1;

  • intsecxtanxdx=secx+csecxtanxdx=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.