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

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Answer 1 · 2015-06-24T16:54:15

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.

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

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

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

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

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

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

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