Marginal Distribution from Bivariate Distribution Matrix

I am doing some practice problems to prepare for my statistics exam, and I just want to know if my reasoning is correct on one problem, and if not, I want to know how I should reason through this. The question is as follows:

Let X and Y follow a bivariate normal distribution with means (3, 2), variances
(1, 4) and covariance c.


$$\begin{bmatrix}X\\Y\end{bmatrix} = N(\begin{bmatrix}3\\2\end{bmatrix}, \begin{bmatrix}1&c\\c&4\end{bmatrix})$$

What is the Marginal Distribution of X?


My initial reasoning is that one method I could go about it is the calculation of the joint PDF and integrating it, but that doesn't sound very clean. A cleaner method could possibly be that since the joint is determined by two normal distributions, then $$f(X,Y) = f(X)f(Y)$$, and $$f_x(X)$$ (the marginal distribution of X) equals $$\int_{-\infty}^{\infty}f(X)f(Y)dY = f(X)\int_{-\infty}^{\infty}f(Y)dY = f(X)$$ since $$f(Y)$$ is a normal distribution, with the final answer being $$f(X) = N(\mu_1, \sigma_1^2) = N(3,1)$$. This seems to make sense but seems a little bit too simplified. Could anyone give me insight on how to tackle this problem, and if I am wrong anywhere, point it out? Thank you.

Bivariate is a special case of jointly normality for 2D, which means these variables are also marginally normal. So, above $$X$$ and $$Y$$ are normal RVs. From the mean vector and the covariance matrix, we know that $$\mu_X=3$$ and $$\sigma^2_X=1$$, so $$X\sim N(3,1)$$. Similarly $$Y\sim N(2,4)$$. So, your final answer is correct.
But, you cannot say $$f(x,y)=f(x)f(y)$$ when $$c\neq 0$$, since it implies independence. Check the bivariate distribution formula here.
And, it is always true that $$\int_y{f(x,y)dy}=f(x)$$, not only when you factorize the joint PDF out, i.e. like $$f(x,y)=f(x)f(y)$$.