Question 1208301
<pre>
Here is the way to solve it by trig substitution. It isn't so bad after all.

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

Draw a right tringle with angle &theta;, hypotenuse 4, opposite leg x, and
adjacent leg {{{sqrt(16-x^2)}}}. As a convention we always put x on the opposite
leg [although it wouldn't matter if we used cofunctions instead.]

[I can't write d&theta; using the program on the site, so I'll use "dO" for "d&theta;", so don't be upset.

{{{drawing(400,240,-.5,4.5,-.5,2.5,

triangle(0,0,4,0,4,2), locate(1.5,-.1,sqrt(16-x^2)), locate(4.1,1,x),
locate(2,1.3,4), locate(.4,.2,theta) )}}}{{{matrix(5,1,

x/4=sin(theta),
x=4sin(theta),
dx=4cos(theta)*"dO",
sqrt(16-x^2)/4=cos(theta),
sqrt(16-x^2)=4cos(theta))}}}

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

Use the substitutions above, taken straight from the right triangle:

{{{int((3(4sin(theta))^3)(4cos(theta))*4cos(theta)dO)}}}

{{{int((3*4^3sin^3(theta)^"")(4cos(theta))*4cos(theta)dO)}}}

Take out all the constants:

{{{3072int(sin^3(theta)cos^2(theta)dO)}}}

When you have an odd power of a sine or cosine, save a sine or cosine
for the derivative and replace the even power that's left with an
identity. so we break up sin<sup>3</sup>&theta; = (sin<sup>2</sup>&theta;)sin&theta; = (1-cos<sup>2</sup>&theta;)sin&theta;

{{{3072int((1-cos^2(theta))*sin^""(theta)cos^2(theta)dO)}}}

{{{3072int(cos^2(theta)sin(theta)dO)}}}{{{""-""}}}{{{3072int(cos^4(theta)sin(theta)dO)}}}

We need to insert a - sign since the derivative of cosine is negative sine:

 {{{-3072int(cos^2(theta)(-sin(theta))dO)}}}{{{""+""}}}{{{3072int(cos^4(theta)(-sin(theta))dO)}}}

{{{-3072(cos^3(theta)/3^"")+3072(cos^5(theta)/5^"")+C}}}

{{{-3072cos^3(theta)( 1/3-expr(1/5)cos^2(theta))+C}}}

{{{-3072(sqrt(16-x^2)/4)^3( 1/3-expr(1/5)(sqrt(16-x^2)/4)^2)+C}}}

{{{-3072((sqrt(16-x^2))^3/64)(1/3-expr(1/5)((16-x^2)/16))+C}}}

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

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

Factor out 1/240

{{{-48(16-x^2)^("3/2")(1/240)(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}}}

That is the same answer as I got above using algebraic substitution.

Edwin</pre>