I got the problem down to
(1/2)y + (1/4)sin2y
the sub I used was x = siny
how do i sub back to X?
I know the first part is sin inverse by how do i get the 2nd part?
If you have $\displaystyle \frac{1}{2}y+\frac{1}{4}sin(2y)$ with $\displaystyle x=sin(y)$, then $\displaystyle y=sin^{-1}x$.
Subbing back in yields $\displaystyle \frac{sin^{-1}x}{2}+\frac{sin(2sin^{-1}x)}{4}$.
Recall the double angle identity: $\displaystyle sin(2\theta)=2sin(\theta)cos(\theta)$.
Letting $\displaystyle \theta= sin^{-1}x$, we have $\displaystyle sin(2sin^{-1}x) = 2sin(sin^{-1}x)cos(sin^{-1}x)$.
Clearly $\displaystyle sin(sin^{-1}x)=x$, but maybe not so clearly: $\displaystyle cos(sin^{-1}x) = \sqrt{1-x^2}$, because if you draw out the triangle, the angle whose sine is $\displaystyle \frac{x}{1}$ will have cosine $\displaystyle \frac{\sqrt{1-x^2}}{1}$.
Thus, $\displaystyle sin(2sin^{-1}x) = 2sin(sin^{-1}x)cos(sin^{-1}x) = 2x\sqrt{1-x^2}.$.
Thus the final answer is $\displaystyle \frac{sin^{-1}x}{2}+\frac{2x\sqrt{1-x^2}}{4} = \frac{1}{2}(sin^{-1}x+x\sqrt{1-x^2}) + C$.
=Thus, $\displaystyle \frac{1}{2}(sin^{-1}x+x\sqrt{1-x^2}) + C = \int \sqrt{1-x^2}\,dx$, which is what I assume you were integrating.
I hope this helped.