Question

How do you integrate `(x^5)(sqrt(4 - x^2)) dx`?

Answer 1

This is of the form:

`sqrt(a^2 - x^2)`

which looks like:

`sqrt(1-1sin^2x) = cosx`

So, let's do the following substitution. Let:

`x = asintheta`
`dx = acosthetad theta`

where `a = sqrt4 = 2`

thus:

`x^5 = 32sin^5theta`
`sqrt(4-x^2) = sqrt(4-4sintheta) = 2costheta`
`dx = 2costhetad theta`

`int x^5sqrt(4-x^2)dx = int 32sin^5theta*4cos^2thetad theta`

`= 128int sin^5thetacos^2thetad theta`

`= 128int (sin^2theta)^2sinthetacos^2thetad theta`

`= 128int (1-cos^2theta)^2cos^2thetasinthetad theta`

Now, let:
`w = costheta`
`dw = -sinthetad theta`

We can then get:

`= -128int (1-w^2)^2w^2dw`

`= -128int (1-2w^2 + w^4)w^2dw`

`= -128int w^2-2w^4 + w^6dw`

`= -128[w^3/3-2/5w^5 + w^7/7]`

Build a triangle; `x/2 = sintheta`, so `sqrt(4-x^2)/2 = costheta`

Thus, we can re-substitute back in the previous values.

`= -128/3cos^3theta + 256/5cos^5theta - 128/7cos^7theta`

`= -cancel(128)^(16)/3(4-x^2)^(3/2)/cancel(8) + cancel(256)^8/5(4-x^2)^(5/2)/cancel(32) - cancel(128)/7(4-x^2)^(7/2)/cancel(128)`

`= -16/3(4-x^2)^(3/2) + 8/5(4-x^2)^(5/2) - 1/7(4-x^2)^(7/2) + C`

Answer 2

Use a `u` substitution to get: `-16/3(4-x^2)^(3/2) +8/5 (4-x^2)^(5/2) - 1/7 (4-x^2)^(7/2) +C`

Explanation:

Although this integral can be evaluated using a trigonometric substitution, it can also be done with a `u` substitution.

`int x^5 sqrt(4-x^2) dx`

Let `u = 4-x^2`, so `du = -2x dx` and we can rewrite the integral:

`int x^5 sqrt(4-x^2) dx = -1/2 int x^4 sqrt(4-x^2) (-2x) dx`

With our choice of `u = 4-x^2`, we also get `x^2 = 4-u`

so `x^4 = (x^2)^2 = (4-u)^2 = 16 - 8u +u^2`

Substituting in the integral yields:

`-1/2 int (16 - 8u +u^2) u^(1/2) du`

Distribute the `u^(1/2)` and integrate term by term.

`-1/2 int (16u^(1/2) - 8u^(3/2) +u^(5/2)) du`

`=-1/2[16 (2/3 u^(3/2)) - 8 (2/5 u^(5/2)) +(2/7u^(7/2))]+C`

Now simplify and rewrite as you see fit.

`= -16/3(4-x^2)^(3/2) +8/5 (4-x^2)^(5/2) - 1/7 (4-x^2)^(7/2) +C`