I have no idea how to obtain the answer to the following problem:
$$I=\int_{0}^{1}\frac{dx}{\sqrt{8-x^2-x^3}}$$
The answer has been given as $$\sin^{-1} \frac{1}{2\sqrt{2}}< I < \frac{1}{\sqrt{2}}\sin^{-1} \frac{1}{2}$$
Any help will be appreciated. Thanks in advance.
Hint: Consider $$\frac{1}{\sqrt{8-x^2}}\leq\frac{1}{\sqrt{8-x^2-x^3}}\leq \frac{1}{\sqrt{8-2x^2}} = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{4-x^2}}$$
$I$ is an elliptic integral, accurate approximations can be deduced from convexity. Over $(0,1)$ we clearly have $8-x^2-x^3 > 8-2x^2$, hence $$ I < \int_{0}^{1}\frac{dx}{\sqrt{8-2x^2}} = \frac{\pi}{6\sqrt{2}} \tag{Ub}$$ but we also have $\frac{1}{\sqrt{8-x^2-x^3}}>\frac{1}{2\sqrt{2}}\left(1+\frac{x^2+x^3}{16}\right)$, hence $$ I > \int_{0}^{1}\frac{1}{2\sqrt{2}}\left(1+\frac{x^2+x^3}{16}\right)\,dx = \frac{199}{384\sqrt{2}}\tag{Lb}$$ and by combining the upper bound and the lower bound $$ 0.3664 < I < 0.3703. \tag{B} $$ $(\text{Lb})$ can be improved through Jensen's inequality: $$ I > \frac{1}{2\sqrt{2}\sqrt{1-\int_{0}^{1}\frac{x^2+x^3}{8}\,dx}}=\sqrt{\frac{12}{89}}>0.36719 $$ and $(\text{Ub})$ can be improved through the Cauchy-Schwarz inequality: $$ I < \sqrt{\int_{0}^{1}\frac{dx}{8-x^2-x^3}}<0.36841 $$ such that $I=\color{green}{0.36}77\pm 0.0008$.
