Definite integration of Bernoulli likelihood Let $q\in(0,1)$ and $x\in\{0,1\}$. The likelihood of a Bernoulli trial is 
$$q^x(1-q)^{(1-x)}.$$
If we take an integration with respect to $q$ over $[0,1]$, the value should be $\frac{1}{2}$: 
$$\int^1_0q^x(1-q)^{(1-x)}dq=B(2,1)=B(1,2)=\frac{1}{2}.$$
My question is, how can we derive the integrated value over $[a,b]$, for some $0\leq a\leq b\leq 1$? 
Would there be closed form solution for the following integral, not using the incomplete beta function?
$$\int^b_aq^x(1-q)^{(1-x)}dq?$$
 A: 
Would there be closed form solution for the following integral, not using the incomplete beta function?

Not really. You won't be able to get away without using special functions.
You can, however, use the hypergeometric function $_2F_1$, but it's really just the incomplete beta in disguise. We set out to evaluate the integral
$$f=\int_0^1 t^{b-1}(1-t)^{c-b-1}(1-xt)^{-a}dt.\qquad \text{Re }c>\text{Re }b>0$$
We have that $$(1-u)^{-s}=\,_1F_0(s;;u)=\sum_{n\ge0}\frac{(s)_n}{n!}u^n$$
where $(x)_n=\frac{\Gamma(x+n)}{\Gamma(x)}$. So then we get
$$\begin{align}
f&=\int_0^1 t^{b-1}(1-t)^{c-b-1}\sum_{n\ge0}\frac{(a)_n}{n!}x^nt^ndt\\
&=\sum_{n\ge0}\frac{(a)_n}{n!}x^n\int_0^1 t^{n+b-1}(1-t)^{c-b-1}dt\\
&=\sum_{n\ge0}\frac{(a)_n}{n!}x^n\frac{\Gamma(n+b)\Gamma(c-b)}{\Gamma(n+c)}\\
&=\frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)}\sum_{n\ge0}\frac{(a)_n}{n!}x^n\frac{\Gamma(n+b)\Gamma(c)}{\Gamma(b)\Gamma(n+c)}\\
&=\mathrm{B}(b,c-b)\sum_{n\ge0}\frac{(a)_n(b)_n}{(c)_n}\frac{x^n}{n!}\\
&=\mathrm{B}(b,c-b)\,_2F_1(a,b;c;x).
\end{align}$$
Then we see that 
$$\begin{align}
\mathrm{B}(x;u,v)&=\int_0^x t^{u-1}(1-t)^{v-1}dt\\
&=x^u\int_0^1 t^{u-1}(1-xt)^{v-1}dt\\
&=x^u\mathrm{B}(u,1)\,_2F_1(1-v,u;1+u;x)\\
&=\frac{x^u}{u}\,_2F_1(u,1-v;1+u;x).
\end{align}$$
Your integral is given by 
$$\int_a^b q^{x}(1-q)^{1-x}dq=\frac{z^{x+1}}{x+1}\,_2F_1(x+1,x-1;x+2;z)\bigg|_{z=a}^{z=b}.$$
