Question 1208301
<pre>

{{{int((3x^3) (sqrt(16-x^2)),dx)}}}

That x<sup>3</sup> factor is going to make trig substitution difficult.
So let's not try trig substitution, but try algebraic substitution instead.

{{{u=16-x^2}}} => {{{x^2=16-u}}}
{{{du=-2x*dx}}}
{{{dx=-du/(2x)}}}

{{{int((3x^3) (sqrt(u))(-du/(2x)))}}}

{{{expr(-3/2)int((x^2) (sqrt(u))du))}}}

{{{expr(-3/2)int((16-u) (sqrt(u))du)))}}}

{{{expr(-3/2)int((16-u) (sqrt(u))du))}}}
{{{expr(-3/2)int((16-u) (u^("1/2")du))}}}
{{{expr(-3/2)int((16u^("1/2")-u^("3/2"))du)}}}
{{{expr(-3/2)(int((16u^("1/2")-u^("3/2"))du))}}}

{{{expr(-3/2)int(16u^("1/2")du+expr(3/2)int(u^("3/2")du^"")))))}}}

{{{-24int(u^("1/2")du+expr(3/2)int(u^("3/2")du^"")))))}}}

{{{-24*(u^("3/2")/expr("3/2")^"")+expr(3/2)(u^("5/2")/expr("5/2")^"")+C}}}

{{{-24*expr(2/3)u^("3/2")+expr(3/2)*expr(2/5)u^("5/2")+C}}}

{{{-16u^("3/2")+expr(3/5)u^("5/2")+C}}}

{{{-16(16-x^2)^("3/2")+expr(3/5)(16-x^2)^("5/2")+C}}}

Your teacher may let you leave it like that, since the
calculus part is over, and the rest is algebra. But
here's the remaining algebra part:

factor out {{{expr(1/5)(16-x^2)^("3/2")}}}

{{{expr(1/5)(16-x^2)^("3/2")(-80+3(16-x^2)^"")+C}}}

{{{expr(1/5)(16-x^2)^("3/2")(-80+48-3x^2)^""+C}}}

{{{expr(1/5)(16-x^2)^("3/2")(-32-3x^2)^""+C}}}

{{{expr(-1/5)(16-x^2)^("3/2")(32+3x^2)^""+C}}}

Edwin</pre>