After using #y=tanx#, #dx=dy/(y^2+1)#, #(sinx)^2=y^2/(y^2+1)# and #(cosx)^2=1/(y^2+1)# transforms, this integral became
#int ((sinx)^8+(cosx)^8)/((sinx)^2*(cosx)^2)*dx#
=#int ((y^2/(y^2+1))^4+(1/(y^2+1))^4)/(y^2/(y^2+1)*1/(y^2+1))*dy/(y^2+1)#
=#int (y^8+1)/[y^2*(y^2+1)^3]*dy#
Now, I decomposed integrand into basic fractions,
#(y^8+1)/[y^2*(y^2+1)^3]#
=#(y^8+3y^6+3y^4+3y+2-3y^6-3y^4-3y^2+1)/[y^2*(y^2+1)^3]#
=#[y^2*(y^2+1)^3]/[y^2*(y^2+1)^3]#-#(3y^6+3y^4+3y^2-1)/[y^2*(y^2+1)^3]#
=#1-(3y^6+3y^4+3y^2-1)/[y^2*(y^2+1)^3]#
=#1-(3y^6+3y^4+3y^2+3-4)/[y^2*(y^2+1)^3]#
=#1-(3y^6+3y^4+3y^2+3)/[y^2*(y^2+1)^3]+4/[y^2*(y^2+1)^3]#
=#1-[3*(y^2+1)(y^4+1)]/[y^2*(y^2+1)^3]+4*[(y^2+1)-y^2]^2/[y^2*(y^2+1)^3]#
=#1-[3*(y^4+1)]/[y^2*(y^2+1)^2]+4*[(y^2+1)^2-2y^2*(y^2+1)+y^4]/[y^2*(y^2+1)^3]#
=#1-[3*(y^4+2y^2+1-2y^2)]/[y^2*(y^2+1)^2]#+#(4*(y^2+1)^2)/[y^2*(y^2+1)^3]#-#(8y^2*(y^2+1))/[y^2*(y^2+1)^3]#+#(4y^4)/[y^2*(y^2+1)^3]#
=#1-[3*(y^2+1)^2]/[y^2*(y^2+1)^2]+(6y^2)/[y^2*(y^2+1)^2]#+#4/[y^2*(y^2+1)]#-#8/(y^2+1)^2#+#(4y^2)/(y^2+1)^3#
=#1-3/y^2+6/(y^2+1)^2#+#4*[(y^2+1)-y^2]/[y^2*(y^2+1)]#-#8/(y^2+1)^2#+#(4y^2+4-4)/(y^2+1)^3#
=#1-3/y^2-2/(y^2+1)^2#+#[4*(y^2+1)]/[y^2*(y^2+1)]-(4y^2)/[y^2*(y^2+1)]#+#(4y^2+4)/(y^2+1)^3-4/(y^2+1)^3#
=#1-3/y^2-2/(y^2+1)^2#+#4/y^2-4/(y^2+1)#+#4/(y^2+1)^2-4/(y^2+1)^3#
=#1+1/y^2+2/(y^2+1)^2-4/(y^2+1)-4/(y^2+1)^3#
Hence,
#(y^8+1)/[y^2*(y^2+1)^3]#
=#int dy+int (dy)/y^2+2int (dy)/(y^2+1)^2-4int (dy)/(y^2+1)-4int (dy)/(y^2+1)^3#
=#y-1/y-4arctany+C+2int (dy)/(y^2+1)^2-4int (dy)/(y^2+1)^3#
#A=2int (dy)/(y^2+1)^2-4int (dy)/(y^2+1)^3#
After using #y=tanz# and #dz=(secz)^2*dz# transform, #A# became
#A=2int ((secz)^2*dz)/(secz)^4-4int ((secz)^2*dz)/(secz)^6#
=#int 2(cosz)^2*dz-int 4(cosz)^4*dz#
=#int 2(cosz)^2*dz-int (2(cosz)^2)^2*dz#
=#int (1+cos2z)*dz-int (1+cos2z)^2*dz#
=-#int cos2z*dz#-#int (cos2z)^2*dz#
=#-1/2sin2z-1/2int (1+cos4z)#
=#-1/2sin2z-1/8sin4z-1/2*z#
=#-1/2*2sinz*cosz-1/8*4sinz*cosz*cos2z-1/2*z#
=#-sinz*cosz-1/2*sinz*cosz*cos2z-1/2*z#
=#-tanz/[(tanz)^2+1]-1/2*tanz/[(tanz)^2+1]*[1-(tanz)^2]/[1+(tanz)^2]-1/2*z#
After using #y=tanz# and #z=arctany# inverse transform,
#A=-y/(y^2+1)-1/2*y*(1-y^2)/(1+y^2)^2-1/2*arctany#
Thus,
#(y^8+1)/[y^2*(y^2+1)^3]#
=#y-1/y-4arctany-y/(y^2+1)-1/2*y*(1-y^2)/(1+y^2)^2-1/2*arctany+C#
=#y-1/y-9/2arctany-y/(y^2+1)-1/2*y*(1-y^2)/(1+y^2)^2+C#
=#tanx-1/tanx-9/2arctan(tanx)-tanx/((tanx)^2+1)-1/2*tanx*(1-(tanx)^2)/(1+(tanx)^2)^2+C#
=#tanx-cotx-(9x)/2-1/2sin2x-1/8sin4x+C#
