How do I find g(x) = ∫(-inf,inf){f(x,y)}dy, given the following f(x,y)

f(x,y) = exp( - (x^2 + y^2 - 2Rxy) / 2(1-R^2) )

(-1
I've already worked out ∫(-inf,inf) ∫(-inf,inf){f(x,y)} dxdy = 2pi for R=0,
and shown that ∫(-inf,inf){exp(-ax^2)} dx = sqrt(pi/a) for R=0 and a>0.
I'm allowed to use these results in my solution.

I think I need to substitute r sinθ or something for y, because f(x,y) is rather complicated, but I don't know what to do when there's a nonzero R, and also as I'm only integrating for y.

Sorry to be asking so many questions, I'd appreciate your help!

You can use a linear u-substitution to complete the square in y. In particular,
(y - Rx)^2 = y^2 - 2Rxy + R^2 x^2

so
x^2 + y^2 - 2Rxy
= y^2 - 2Rxy + R^2 x^2 - R^2 x^2 + x^2
= (y - Rx)^2 + x^2 (1-R^2)

Plugging this in to your function gives
f(x, y) = exp( - ( (y-Rx)^2 + x^2 (1-R^2) ) / 2(1-R^2))
= exp(-x^2 / 2) * exp(-(y-Rx)^2 / 2(1-R^2))

From here use u = y-Rx and your previous result.

fyi what you are doing is working on the bivariate normal probability distribution. In this case the variables X and Y have mean 0, std deviation 1 and correlation R. Your previous results confirm that the integral over the whole x-y space is 1 (well you got 2pi, but this just means the probability needs to be normalized by dividing by 2pi). The question is asking for the conditional distribution of X given Y, a quantity that comes in useful in many applications.