Question

How do you find the definite integral of `(x^3)/sqrt(16 - x^2)` in the interval `[0, 2sqrt3]`?

Answer 1

`40/3`.

Explanation:

Let, `I=int_0^(2sqrt3)x^3/sqrt(16-x^2)dx`.

We subst. `16-x^2=t^2. :. x^2=16-t^2`.

`:. 2xdx=-2tdt, or, xdx=-tdt`.

Also, when `x=0, t=4, &, x=2sqrt3, t=2`.

`:. I=int_0^(2sqrt3)x^3/sqrt(16-x^2)dx`,

`=int_0^(2sqrt3)x^2/sqrt(16-x^2)xdx`,

`=int_4^2(16-t^2)/sqrt(t^2)(-tdt)`,

`=int_4^2(t^2-16)dt`,

`=[t^3/3-16t]_4^2`,

`=[{8/3-32}-{64/3-64}]`,

`=32-56/3`.

`rArr I=40/3`, as desired!

Answer 2

`I=40/3`

Explanation:

Here,

`I=int_0^(2sqrt3) x^3/sqrt(16-x^2)dx`

Subst. `x=4sinu=>dx=4cosudu`

`x=0=>4sinu=0=>sinu=0=>u=0`

`x=2sqrt3=>4sinu=2sqrt3=>sinu=sqrt3/2=>u=pi/3`

So,

`I=int_0^(pi/3)(64sin^3u)/sqrt(16-16sin^2u)xx4cosudu`

`=int_0^(pi/3)(64sin^3u)/(4cosu)xx4cosudu`

`=int_0^(pi/3)64sin^3udu`

`=16int_0^(pi/3) color(red)(4sin^3u)du`

`=16int_0^(pi/3)color(red)((3sinu-sin3u))du`

`=16[-3cosu+(cos3u)/3]_0^(pi/3)`

`=16{[-3(1/2)+(-1)/3]-[-3(1)+1/3]}`

`=16{-3/2-1/3+3-1/3}`

`=16{3-3/2-1/3-1/3}`

`=16{(18-9-2-2)/6}`

`=16xx5/6`

`=8xx5/3`

`=40/3`

NOTE :

`color(red)(sin3theta=3sintheta-4sin^3theta`

`color(red)(4sin^3theta=3sintheta-sin3theta`

Answer 3

Make a trig substitution to simplify the denominator, then evaluate as normal

Explanation:

Note first that the function has two values at which it blows up, so can't be integrated over: `x=+-4`. The given limits of integration lie between these values, so this is no problem.

Substitute `x=4sin theta` to turn the denominator into a simple trig function via the identity `sin^2 theta +cos^2 theta=1`. Note that `(dx)/(d theta)=4 cos theta` and that the limits of integration `x=0,2sqrt(3)` are equivalent to `theta=0,pi/3`.
(Recall that `sin (pi/3) = sqrt(3)/2`)

`int_0^(2sqrt(3)) x^3/sqrt(16-x^2)dx=int_0^(pi/3) (64sin^3theta)/sqrt(16-16sin^2theta)*4cos theta d theta`

Simplify:
`64int_0^(pi/3) sin^3theta d theta`

Use the same identity to rewrite this as
`64int_0^(pi/3) (1-cos^2theta)sin theta d theta=64int_0^(pi/3) sin theta-sin theta cos^2theta d theta`

This integrates readily as
`64[-cos theta+1/3cos^3theta)]_(theta=0)^(pi/3)`

The limits of integration give us now
`64[(-1/2+1/3*1/8)-(-1+1/3*1)]`
`=40/3`