Given $\displaystyle (sin^{-1}x)^2=x^2+\frac{1}{3}x^4+...$, use the standard series expansions for ln(1+x) and $\displaystyle e^x$ to show that when $\displaystyle x$ is small enough for powers of x above fourth to be neglected, $\displaystyle (sin^{-1}x)^2=\frac{1}{6}[ln(1+x^2)+5e^{x^2}-5]$