How do you integrate #int 2x^5sqrt(2+9x^2)# using trig substitutions?
1 Answer
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" "" "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
#=1/729(2/7u^(7/2)-8/5u^(5/2)+8/3u^(3/2))+C#
Returning to
#=(2(2+9x^2)^(7/2))/5103-(8(2+9x^2)^(3/2))/3645+(8(2+9x^2)^(3/2))/2187+C#