Answer
(a) $\frac{1}{\sqrt {1-x^{2}}}=1+\Sigma_{n=1}^{\infty}\frac{1.3.5.7......(2n-1)x^{2n}}{2^{n}n!}$
(b) $sin^{-1}x=x+\Sigma_{n=1}^{\infty}\frac{1.3.5.7......(2n-1)x^{2n+1}}{(2n+1)2^{n}n!}$
Work Step by Step
(a) $\frac{1}{\sqrt {1-x^{2}}}=(1-x^{2})^{-1/2}=(1+(-x^{2}))^{-1/2}$
$(1+(-x^{2}))^{-1/2}=\Sigma_{n=1}^{\infty}(-1/2)n(-1)^{n}x^{2n}$
$\frac{1}{\sqrt {1-x^{2}}}=1+\Sigma_{n=1}^{\infty}\frac{1.3.5.7......(2n-1)x^{2n}}{2^{n}n!}$
(b) From part (a), we have
$\frac{1}{\sqrt {1-x^{2}}}=1+\Sigma_{n=1}^{\infty}\frac{1.3.5.7......(2n-1)x^{2n}}{2^{n}n!}$
The derivative of $sin^{-1}x$ is $\frac{1}{\sqrt {1-x^{2}}}$
$\int (1+\Sigma_{n=1}^{\infty}\frac{1.3.5.7......(2n-1)x^{2n}}{2^{n}n!})dx=c+x+\Sigma_{n=1}^{\infty}\frac{1.3.5.7......(2n-1)x^{2n+1}}{(2n+1)2^{n}n!}$
Set $x=0$ to find $c$
$sin^{-1}(0)=c+0+0$
$c=0$
Thus, $sin^{-1}x=x+\Sigma_{n=1}^{\infty}\frac{1.3.5.7......(2n-1)x^{2n+1}}{(2n+1)2^{n}n!}$
