Using a simplified notation, you have $\textit{a priori}$ that $\theta\sim Beta(a,b)$, and $$ f(x\mid\theta)={n\choose x}\theta^x (1-\theta)^{n-x} \, , $$ for $x=0,1,\dots,n$ and $\theta\in[0,1]$.

As you already know, the posterior $\pi(\theta\mid x)$ is proportional to $$ Likelihood \times Prior \, , $$ where the $Likelihood$ is just $f(x\mid\theta)$ seen as function of $\theta$, for some fixed value of the observation $x$.

The important point here, which happens over and over in this kind of computation, is that you don't need to evaluate the integral $\int_0^1\pi(\theta\mid x)\,d\theta$ to find the normalization constant. For example, if the posterior is proportional to $$ \theta^5 (1-\theta)^7 = \theta^{6-1} (1-\theta)^{8-1} \, , $$ then you already know, by simple inspection of the formula of the beta density, that $\theta\mid x\sim Beta(6,8)$.

Can you see why?

This is a pattern that you should keep in your mind when solving similar problems.

Using a simplified notation, you have $\textit{a priori}$ that $\theta\sim Beta(a,b)$, and $$ f(x\mid\theta)={n\choose x}\theta^x (1-\theta)^{n-x} \, , $$ for $x=0,1,\dots,n$ and $\theta\in[0,1]$.

As you already know, the posterior $\pi(\theta\mid x)$ is proportional to $$ Likelihood \times Prior \, , $$ where the $Likelihood$ is just $f(x\mid\theta)$ seen as function of $\theta$, for some fixed value of the observation $x$.

The important point here, which happens over and over in this kind of computation, is that you don't need to evaluate the integral $\int_0^1\pi(\theta\mid x)\,d\theta$ to find the normalization constant of the posterior density. For example, if the posterior is proportional to $$ \theta^5 (1-\theta)^7 = \theta^{6-1} (1-\theta)^{8-1} \, , $$ then you already know, by simple inspection of the formula of the beta density, that $\theta\mid x\sim Beta(6,8)$.

Can you see why?

This is a pattern that you should keep in your mind when solving similar problems.