Question 276711
{{{1/sqrt((4-3x^2))^3}}}
<pre><font size = 4 color = "indigo"><b>
Express the denominator as the {{{3/2}}} power

{{{1/((4-3x^2)^(3/2))}}}

Then as a negative exponent:

{{{(4-3x^2)^(-3/2)}}}

We want to get a binomial that starts with 1, not 4.

Factor out 4 in the parentheses:

{{{(4(1-(3/4)x^2))^(-3/2)}}}

{{{4^(-3/2)(1-3/4)^(-3/2)}}}


{{{4^(-3/2)}}} simplifies to {{{1/8}}}


So we have

{{{(1/8)(1-(3/4)x^2)^(-3/2)}}}

So we now expand 

{{{(1-(3/4)x^2)^(-3/2)}}} and then multiply it by {{{1/8}}}

We use the infinite binomial series:

{{{(1+u)^n=1+nx+((n(n-1))/2!)u^2 + ((n(n-1)(n-2))/3!)u^3+"..."}}}

Substituting {{{n=-3/2}}},  {{{((n(n-1))/2!)}}} becomes {{{15/8}}}

Substituting {{{u=(3/4)x^2}}}, {{{u^2}}} becomes {{{(9/16)x^4}}}

Therefore {{{((n(n-1))/2!)u^2}}} becomes {{{(15/8)(9/16)x^4}}} or {{{(135/128)x^4}}}


Substituting {{{n=-3/2}}}, {{{(n(n-1)(n-2))/3!}}} becomes {{{-35/16}}}

Substituting {{{u=(3/4)x^2}}}, {{{u^3}}} becomes {{{(27/64)x^6}}}

Therefore {{{( (n(n-1)(n-2)) /3!)u^3}}} becomes {{{(-35/16)(27/64)x^6}}} or

{{{(-945/1024)x^6}}}

So the expansion to 4 terms of {{{(1-(3/4)x^2)^(-3/2)}}}

is

{{{1-(3/2)x^2+(15/8)x^4-(945/1024)x^6+"..."}}}

So we multiply that by {{{1/8}}} and that will be the binomial
series to 4 terms.

Edwin</pre>