Question

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

Answer 1

`(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`