How do you integrate #intx^3sqrt(16 - x^2) dx#?

1 Answer
Jun 24, 2015

I would let:
#x = 4sintheta#
#dx = 4costhetad theta#
#sqrt(16 - x^2) = 4costheta#
#x^3 = 64sin^3theta#

#=>1024int sin^3thetacos^2thetad theta#

#=1024int sin^2thetasinthetacos^2thetad theta#

#=1024int (1-cos^2theta)cos^2thetasinthetad theta#

At this point you can let:
#u = costheta#
#du = -sinthetad theta#

#=> 1024int (cos^2theta - 1)cos^2theta(-sintheta)d theta#

#= 1024int (u^2 - 1)u^2du#

#= 1024int u^4 - u^2du#

#= 1024(u^5/5 - u^3/3)#

#= 1024(cos^5theta/5 - cos^3theta/3)#

Since #costheta = sqrt(16-x^2)/4#:

#= 1024((sqrt(16-x^2)/4)^5/5 - (sqrt(16-x^2)/4)^3/3)#

#= 1024(((16-x^2)^(5/2)/1024)/5 - ((16-x^2)^(3/2)/64)/3)#

#= 1024((16-x^2)^(5/2)/(5*1024) - (16-x^2)^(3/2)/(3*64))#

#= cancel(1024)((16-x^2)^(5/2)/(5*cancel(1024)) - (16-x^2)^(3/2)/(3/16*cancel(1024)))#

#= color(green)(1/5(16-x^2)^("5/2") - 16/3(16-x^2)^("3/2") + C)#

Normally here would be okay, but we can go a bit further.

#= (16-x^2)^(3/2)(1/5(16-x^2) - 16/3)#

#= (16-x^2)^(3/2)(3/15(16-x^2) - 80/15)#

#= 1/15(16-x^2)^(3/2)(3(16-x^2) - 80)#

#= 1/15(16-x^2)^(3/2)(48-3x^2 - 80)#

#= 1/15(16-x^2)^(3/2)(-32-3x^2)#

#= color(blue)(-1/15(16-x^2)^("3/2")(3x^2 + 32) + C)#