Suppose I have a random variable $x$, and I want to perform the integral of a function of $x$ such that: $$y=f(x)=\int_{c_l}^{c_u} f(x,c) dc$$ where $f(x,c)$ is a nonlinear function of $x$ and $c$. Definitely $y$ is going to be random variable dependent on $x$. If $y$ does not have a closed form expression is there a way to find the distribution of $y$, given distribution of $x$.
I have studied a little about transformation of random variables, but there integrals are not considered.
Though the question has been answered. I will be grateful if one can provide me some literature regarding same. This will help me in knowing things better.
Let us say $Y = \int_0^1 f(X, t) dt$ (since $c_u, c_l$ are fixed, we may as well assume they are $0$ and $1$ for simplicity). I will denote $I(x) := \int_0^1 f(x, t) dt.$ To compute the desired density, we will make a rather strong assumption, but as I showed in this other MSE post we cannot say a lot more in the general setting. Thus we will make the assumption that $I$ is invertible and differentiable with respect to $x$ (implicitly here we make assumptions on $f(\cdot, \cdot),$ which may for example be consulted here). Note that the density of a random variable $X$ is nothing more than the Radon-Nikodym derivative of the pushforward measure $X_* \mathbb{P}$ of $X$ with respect to the Lebesgue measure $\lambda,$ where we have denoted by $\mathbb{P}$ the measure on the domain of the random variable. The following discussion will rely on the notions mentioned above, so I do urge you to familiarise yourself with them.
We have that $Y = I(X)$ and we want to compute the Radon-Nikodym derivative $\frac{d I_* X_* \mathbb{P}}{d \lambda}.$ Since we assumed the necessary conditions to make $I$ differentiable, we have that $I'(x) = \int_0^1 \frac{\partial f}{\partial x} (x, t) dt$. Let us also denote the density of $X$ by $g.$ It follows by basic results concerning Radon-Nikodym derivatives and the ones linked above that: $$d I_* X_* \mathbb{P} = \frac{d I_* X_* \mathbb{P}}{d I_* \lambda} \frac{d I_* \lambda}{d \lambda} d \lambda = \frac{d X_* \mathbb{P}}{d \lambda} \circ I^{-1} (I^{-1})' d \lambda,$$ so the density of $Y$ is given by $$ \rho(x) = g(I^{-1}(x)) \frac{1}{\int_0^1 \frac{\partial f}{\partial x} (I^{-1}(x), t) dt}.$$ I hope this helps. :)