How do you integrate #int 2x^5sqrt(2+9x^2)# using trig substitutions?

1 Answer
Feb 2, 2017

#(2(2+9x^2)^(7/2))/5103-(8(2+9x^2)^(3/2))/3645+(8(2+9x^2)^(3/2))/2187+C#

Explanation:

#int2x^5sqrt(2+9x^2)color(white).dx#

The easiest way to do this is to, in fact, use the substitution #u=2+9x^2#. Note that this can be manipulated to show the following things. (The ones with stars will be substituted into the integral.)

  • #u=2+9x^2" "" "star#
  • #u-2=9x^2#
  • #x^2=(u-2)/9#
  • #x^4=(u-2)^2/81" "" "star#
  • #du=18xcolor(white).dx#
  • #1/18du=xcolor(white).dx" "" "star#

The integral can be rewritten:

#=int2x^4sqrt(2+9x^2)(xcolor(white).dx)=int2(u-2)^2/81sqrtu(1/18du)#

This simplifies:

#=1/729int(u-2)^2sqrtucolor(white).du#

Expanding and distributing:

#=1/729int(u^2-4u+4)(u^(1/2))color(white).du#

#=1/729int(u^(5/2)-4u^(3/2)+4u^(1/2))du#

Using #intu^ncolor(white).du=u^(n+1)/(n+1)+C#:

#=1/729(2/7u^(7/2)-8/5u^(5/2)+8/3u^(3/2))+C#

Returning to #x# from #u=2+9x^2#:

#=(2(2+9x^2)^(7/2))/5103-(8(2+9x^2)^(3/2))/3645+(8(2+9x^2)^(3/2))/2187+C#