Why is the posterior distribution in Bayesian Inference often intractable? I have a problem understanding why Bayesian Inference leads to intractable problems. The problem is often explained like this:

What I don't understand is why this integral has to be evaluated in the first place: It seems to me that the result of the integral is simply a normalization constant (as the dataset D is given). Why can one not simply calculate the posterior distribution as the numerator of the right-hand side and then infer this normalization constant by requiring that the integral over the posterior distribution has to be 1?
What am I missing?
Thanks!
 A: 
Why can one not simply calculate the posterior distribution as the numerator of the right-hand side and then infer this normalization constant by requiring that the integral over the posterior distribution has to be 1?

This is precisely what is being done. The posterior distribution is
$$P(\theta|D) = \dfrac{p(D|\theta) \, P(\theta)}{P(D)}. $$
The numerator on the right hand side is $P(D|\theta)P(\theta)$. This is a function over $\theta$ and to be a probability distribution, it has to integrate to 1. Thus we need to find the constant $c$, such that 
\begin{align*}
&\int_{\theta} cP(D|\theta) \, P(\theta)\, d\theta = 1\\
\Rightarrow & \int_{\theta} cP(D, \theta) \, d\theta = 1\\
\Rightarrow & cP(D) = 1\\
\Rightarrow& c = \dfrac{1}{P(D)}.
\end{align*}
Thus, the normalizing constant is $P(D)$ which is often intractable, or overtly complicated.
A: I had the same question. This great post explains it really well.
In a nutshell. It is intractable because the denominator has to evaluate the probability for ALL possible values of ; in most interesting cases ALL is a large amount. Whereas the numerator is for a single realization of .   
See Eqs. 4-8 in the post. Screenshot of the link:

