1. ## Trig Substitution Help

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?

2. If you have $\frac{1}{2}y+\frac{1}{4}sin(2y)$ with $x=sin(y)$, then $y=sin^{-1}x$.

Subbing back in yields $\frac{sin^{-1}x}{2}+\frac{sin(2sin^{-1}x)}{4}$.

Recall the double angle identity: $sin(2\theta)=2sin(\theta)cos(\theta)$.

Letting $\theta= sin^{-1}x$, we have $sin(2sin^{-1}x) = 2sin(sin^{-1}x)cos(sin^{-1}x)$.

Clearly $sin(sin^{-1}x)=x$, but maybe not so clearly: $cos(sin^{-1}x) = \sqrt{1-x^2}$, because if you draw out the triangle, the angle whose sine is $\frac{x}{1}$ will have cosine $\frac{\sqrt{1-x^2}}{1}$.

Thus, $sin(2sin^{-1}x) = 2sin(sin^{-1}x)cos(sin^{-1}x) = 2x\sqrt{1-x^2}.$.

Thus the final answer is $\frac{sin^{-1}x}{2}+\frac{2x\sqrt{1-x^2}}{4} = \frac{1}{2}(sin^{-1}x+x\sqrt{1-x^2}) + C$.

=Thus, $\frac{1}{2}(sin^{-1}x+x\sqrt{1-x^2}) + C = \int \sqrt{1-x^2}\,dx$, which is what I assume you were integrating.

I hope this helped.

3. Yeah I got it. Thanks.