Is there such a thing as implicit integration? A problem I'm working on asks me at a certain point to integrate the following:
$$x=2\left(\sin\left(y\right)\right)^2.$$
I've never integrated anything in this form. I can't isolate for $y$ either. Is there some sort of implicit integration technique I should be aware of, to get the indefinite integral of this?
Thanks to the below comments, I was able to isolate y and get
$$\sin^{-1}\left(\sqrt{\left(\frac{x}{2}\right)}\right)=y$$
....which still looks insanely messy to integrate. I imagine I can use a u-substitution or something similar to solve this? 
 A: The question is from a very long time ago, but I can still attempt to provide an answer. Implicit integration is kind of like the topic in differential equations called exact differential equations. It’s pretty much tracing backwards from applying multivariable chain rule on a function of multiple variables. I’d say it’s the closest thing I’ve seen to a concept of “implicit integration”.
A: Well somewhat, my friend and I derived a method that can be somewhat determined as a form of "implicit integration".  Where by multiplying $y'$ to both sides of an equation it can easily be manipulated and integrated to get an answer.
First of we make the equation as easy as possible to integrate on both sides, i.e:
$\sin^2(y)=\frac{x}{2}$
Next we multiply $y'$ to both sides you'll see why this is useful in a second.
$y'\sin^2(y)=\frac{xy'}{2}$
Then we just integrate both sides of this equation with respect to x, treating y as a function of x.
$\int y'\sin^2(y)dx =\int\frac{xy'}{2}dx$
We solve the left by using a neat little Lemma which you can prove with u-sub:
$\int y'f(x) dx=\int f(y)dy$
We can solve the left using an IBP to achieve an answer of $\int\frac{xy'}{2}=\frac{1}{2}(xy-Y)$ where $Y$ is the integral of y, which we need to solve for.
So we get:
$\int \sin^2(y) dy = \frac{1}{2}(xy-Y)$ After integrating $\sin^2(y)$ we get:
$\frac{1}{2}(y-\frac{1}{2}\sin(2y)) = \frac{1}{2}(xy-Y)$ $\frac{1}{2}$'s cancel out and solving for $Y$ yields the following:
$Y = xy-(y-\frac{1}{2}\sin(2y))$ then substituting $y=\arcsin(\sqrt{\frac{x}{2}})$
We arrive at our final integral, which would be the answer if you had integrated normally, notably:
$Y = \frac{1}{2}\sqrt{-x(x-2)}+(x-1)\arcsin\left(\sqrt{\frac{x}{2}}\right)$
Note: Also I know I'm a little late and this response may never be seen but here it is anyway :)
