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

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#

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#