Integral of Sec^2 x/(secax+tanx) ^9/2 dx=?

1 Answer
Mar 19, 2018

Answer:

#-1/7(secx+tanx)^(-7/2)-1/11(secx+tanx)^(-11/2)+C#

Explanation:

#int (secx)^2/(secx+tanx)^(9/2)*dx#

=#1/2int (2(secx)^2)/(secx+tanx)^(9/2)*dx#

=#1/2int secx*(secx+tanx)/(secx+tanx)^(9/2)dx#+#1/2int secx*(secx-tanx)/(secx+tanx)^(9/2)dx#

=#1/2int [(secx)^2+secx*tanx]/(secx+tanx)^(9/2)dx#+#1/2int secx*[(secx)^2-(tanx)^2]/(secx+tanx)^(11/2)dx#

=#1/2int [(secx)^2+secx*tanx]/(secx+tanx)^(9/2)dx#+#1/2int secx/(secx+tanx)^(11/2)dx#

=#1/2int [(secx)^2+secx*tanx]/(secx+tanx)^(9/2)dx#+#1/2int [(secx)^2+secx*tanx]/(secx+tanx)^(13/2)dx#

After using #y=secx+tanx# and #dy=[(secx)^2+secx*tanx]*dx# transforms, this integral became

#1/2int (dy)/y^(9/2)#+#1/2int (dy)/y^(13/2)dx#

=#-1/7y^(-7/2)-1/11y^(-11/2)+C#

=#-1/7(secx+tanx)^(-7/2)-1/11(secx+tanx)^(-11/2)+C#