Integrating $\int^b_a(x-a)^3(b-x)^4 \,dx$ I came across a question today...

The value of $\int^b_a(x-a)^3(b-x)^4 \,dx$ is

First I tried the property $\int^b_af(x)=\int^b_af(a+b-x)$. I got $\int^b_a(x-a)^4(b-x)^3 \,dx$, which can be simplified to: $\dfrac{b-a}{2}\int^b_a(x-a)^3(b-x)^3 \,dx$. Well now what? Do I have to open these brackets and do the whole thing or is there a short way (as definite integrals always have)?

The Beta function integral for integer arguments is proven using integration by parts. That having been done, we get
$$
\begin{align}
\int_a^b(x-a)^3(b-x)^4\,\mathrm{d}x
&=\int_0^{b-a}x^3(b-a-x)^4\,\mathrm{d}x\tag{1}\\
&=(b-a)^8\int_0^1x^3(1-x)^4\,\mathrm{d}x\tag{2}\\[3pt]
&=(b-a)^8\mathrm{B}(4,5)\tag{3}\\[6pt]
&=(b-a)^8\frac{\Gamma(4)\Gamma(5)}{\Gamma(9)}\tag{4}\\
&=(b-a)^8\frac{3!\,4!}{8!}\tag{5}\\[3pt]
&=\frac{(b-a)^8}{280}\tag{6}
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x+a$
$(2)$: substitute $x\mapsto (b-a)x$
$(3)$: apply Beta function
$(4)$: convert to Gamma function
$(5)$: convert to factorials
$(6)$: evaluate
