Maclaurin series?

Question

a) use the binomial series to expand 1/sqrt of (1-x^2)
b) use part (a) to find the Maclaurin series for sin^(-1)x

show work thanks

Answer 1

(1-x^2)^(-1/2)
= 1 + (-x^2) (-1/2) + (-x^2)^2 (-1/2) (-3/2) / 2! + (-x^2)^3 (-1/2) (-3/2) (-5/2) / 3! + ...
= 1 + (x^2 / 2) + 3 (x^4 / 4) / 2! + 15 (x^6 / 8) / 3! + ... + (2n-1)!! (x^2 / 2)^n / n! + ...
where !! is the double factorial, e.g. 9!! = 9.7.5.3.1

Now d/dx arcsin x = (1-x^2)^(-1/2), so the Maclaurin series for arcsin x is just the integral of this one. Note that arcsin 0 = 0, so the constant of integration will be 0. So we get
arcsin x = x + x^3 / 6 + 3 x^5 / 40 + ... + (2n-1)!! x^(2n+1) / [(2^n (2n+1) n!] + ...

Answer 2

a) 1/sqrt of (1-x^2) = (1-x^2)^(-1/2)

(1-x^2)^(-1/2)
= 1 + [(-1/2)/1] (-x^2) + [(-1/2)(-3/2)/2!] (-x^2)^2 +
[(-1/2)(-3/2)(-5/2)/3!] (-x^2)^3 +
[(-1/2)(-3/2)(-5/2)(-7/2)/4!] (-x^2)^4 + ...
= 1 + (1/2)(x^2) + (3/8)(x^4) + (5/16)(x^6) + ...

b) Due to d/dx [arcsin x] = (1-x^2)^(-1/2)

So,
arcsin x = Int [(1-x^2)^(-1/2)] dx
arcsin x = Int [1+(1/2)(x^2)+(3/8)(x^4)+(5/16)(x^6)+...] dx
arcsin x = x+(1/6)(x^3)+(3/40)(x^5)+(5/112)x^7+...

You may check the formula of arcsin x at below link:
http://en.wikipedia.org/wiki/Maclaurin_series