Why do we need to a canonical transformation? Is it to simplify the form of Hamilton's equation?
It is just for our convenience. Mathematically, they preserve the Hamiltonian structure i.e. all the different forms represent the same dynamical system. When there are several choices to describe the system under consideration, one set of equations may be convenient over the rest. We tend to express our equations such that we find a maximum number of cyclic co-ordinates.
For example, to discuss the motion of a particle in a plane under the influence of a central force, we may use generalized co-ordinates as
Cartesian co-ordinates: $\quad q_1 = x, \quad q_2=y \quad$ both of which are not cyclic.
Plane polar co-ordinates: $\quad q_1 = r, \quad q_2=\theta \quad$ in which $\theta$ is cyclic.
The number of cyclic co-ordinates can thus depend on the choice of generalized co-ordinates. For each problem, there are particular choices in which many or even all co-ordinates are cyclic. Once we find the set, the calculations are trivial.
Why is it trivial?
Suppose there exist a system with all its co-ordinates $q_i$ cyclic and $H$ is a constant of motion. Then, the conjugate momenta $p_i$ are all constants $\alpha_i$ since, $\dot{p_i} = -\dfrac{\partial H}{\partial q_i}=0$. So Hamilton's equations of motion would give
$$\dot{q_i} = \dfrac{\partial H}{\partial \alpha_i} = \beta_i$$ with each $\beta_i$ is a function of only $\alpha_i$ and are independent of time. Hence, $q_i = \beta_i t + \delta_i$ as easier as it could get.