Solution
= Solution
With the <change of variables> $u=\tan x$, one has $dx=du/(1+u^2)$, $\sin x=u/\sqrt{1+u^2}$ and $\cos x=1/\sqrt{1+u^2}$. The transformed <improper integral> is integrable at its zero endpoint and becomes
$$
\int_0^{1/\sqrt3}(u^{-1/2}+u^{3/2})du=\left[2u^{1/2}+\frac25u^{5/2}\right]_0^{1/\sqrt3}.
$$
Hence
$$
\boxed{I=\frac{32}{15\sqrt[4]3}.}
$$