In fact, we can reduce it to an elliptic integral.
To see this, let u = sin x, x = sin-1u,
dx = du/root(1-u²)
Then the integral becomes
Integration (root(u) du/root(1-u²)),
and, rationalising the denominator,
Integr (root(u-u³) du/(1-u²))
Since you have the square root of a cubic polynomial
in the integrand, you now have an elliptic integral.
Finally, here is the result from the Wolfram
integrator , a theorem used in Engineering Mathematics for the original problem:
Root sinx cannot be integrated in usual manner ie it's non-integrable.It is an elliptic integral.