Substitute #x =1/2 sect# with #t in (0,pi/2)#, #dx = 1/2 sect tant dt#:
#int sqrt(20x^2-5)dx = 1/2 int sqrt(20( sect/2)^2-5)sect tant dt#
#int sqrt(20x^2-5)dx = 1/2 int sqrt(5 sec^2t-5)sect tant dt#
#int sqrt(20x^2-5)dx = sqrt5/2 int sqrt( sec^2t-1)sect tant dt#
Use now the trigonometric identity:
#sec^2t-1 = tant^2t#
and as #tanx > 0# for #t in (0,pi/2)#:
#sqrt(sec^2t-1) = tant#
Then:
#int sqrt(20x^2-5)dx = sqrt5/2 int sect tan^2t dt#
and using the same identity:
#int sqrt(20x^2-5)dx = sqrt5/2 int sect (sec^2t-1) dt#
#int sqrt(20x^2-5)dx = sqrt5/2 int sec^3tdt - sqrt5/2 int sectdt #
Solve now the integrals:
#int sect = ln abs (sect+tant)+C#
and:
#int sec^3tdt = int sectd/dt(tant)dt#
Integrate by parts:
#int sec^3tdt = sect tant - int tant d/dt(sect)dt#
#int sec^3tdt = sect tant - int sect tan^2t dt#
#int sec^3tdt = sect tant - int sect (sec^2t-1) dt#
#int sec^3tdt = sect tant - int sec^3tdt +int sectdt #
Solve now for the original integral:
#2int sec^3tdt = sect tant +int sectdt #
#int sec^3tdt = (sect tant)/2 +1/2 int sectdt #
#int sec^3tdt = (sect tant)/2 +1/2 ln abs (sect+tant) +C#
Then:
#int sqrt(20x^2-5)dx = sqrt5/4 sect tant - sqrt5/4 ln abs (sect+tant) +C #
Undo the substitution:
#tant = sqrt(sec^2t-1) = sqrt(4x^2-1) #
#sect = 2x#
so:
#int sqrt(20x^2-5)dx = sqrt5/2 xsqrt(4x^2-1) - sqrt5/4 ln (2absx+sqrt(4x^2-1)) +C #
