How do I evaluate int\sqrt{20x^{2}-5}dx?

1 Answer
Jun 28, 2018

int sqrt(20x^2-5)dx = sqrt5/2 xsqrt(4x^2-1) - sqrt5/4 ln (2absx+sqrt(4x^2-1)) +C

Explanation:

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