Question 445411
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{{{int( (x*sqrt(1-x^2) dx))}}}

Rewrite as

{{{int((  (1-x^2)^(1/2)x*dx  ))}}}

Let {{{u=1-x^2}}}

Then {{{du=-2x*dx}}}

Solve for dx.  {{{dx=(du)/(-2x)}}}

Substitute for u and du

{{{int((  (1-x^2)^(1/2)x*dx  ))}}}

{{{int((  u^(1/2)x*expr(du/(-2x))  ))}}}

Cancel the x's

{{{int((  u^(1/2)cross(x)*expr(du/(-2cross(x)))  ))}}}


The -2 on the bottom is the same as a multiplication by
{{{-1/2}}} and can be brought out in front of the integral:

{{{expr(-1/2)int((  u^(1/2)du)  ))}}}

Next we use the formula {{{int(u^n*du)}}}{{{""=""}}}{{{expr(u^(n+1)/(n+1))+C}}}
with {{{n=1/2}}}

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

The {{{3/2}}} on the bottom is a division by {{{3/2}}} which is the
same as a multiplication by {{{2/3}}} which we can take out in front
of the integral sign.  So we have:

{{{expr(2/3)*expr(-1/2)(u^(3/2))+C}}}{{{""=""}}}{{{expr(-1/3)(u^(3/2))+C}}}

Finally we replace u using {{{u=1-x^2}}}

{{{expr(-1/3)(1-x^2)^(3/2)+C}}}

-----------------------

That's a long process, but later you will learn this shortcut:

{{{int((  (1-x^2)^(1/2)x*dx  ))}}}

Since the derivative of {{{1-x^2}}} is {{{-2x}}}, and we
already have an x factor, we only need to put in {{{-2}}}
and take out {{{-1/2}}}, since they would cancel:

{{{red(expr(-1/2))int((  (1-x^2)^(1/2)red(-2)xdx  ))}}}

Then you just observe that that is in the form {{{int(u^n*du)}}}
without writing any u's or du's.

Then you add 1 to the exponent and multiply by the reciprocal
of the new exponent:

{{{expr(-1/2)expr(2/3)(1-x^2)^(3/2)+C}}}
 
{{{expr(-1/3)(1-x^2)^(3/2)+C}}}

Edwin</pre>