After your trig. substitution and appliction of the double-angle identity for cosine, you should have:
$$I=\frac{1}{2}\int 1+\cos(2\theta)\,d\theta=\frac{1}{2}\left(\theta+\frac{1}{2}\sin(2\theta)\right)+C$$
Okay, now at this point we can apply a double-angle identity for sine and state:
$$I=\frac{1}{2}\left(\theta+\sin(\theta)\cos(\theta)\right)+C$$
Okay, now you can back-substitute for $\theta$, observing that:
$$\theta=\arcsin(y)$$
$$\sin(\theta)=y$$
$$\cos(\theta)=\sqrt{1-y^2}$$
And so we find:
$$I=\frac{1}{2}\left(\arcsin(y)+y\sqrt{1-y^2}\right)+C$$
And that's the final answer...no need to "get rid" of the inverse sine function. :D
