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)?
 A: 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
A: How about the following way? (though I'm not sure if you like it)
$$\begin{align}\\&\int_{a}^{b}(\color{red}{x-a})^3(b-x)^4dx\\&=\int_{a}^{b}(\color{red}{x-b+b-a})^3(x-b)^4dx\\&=\int_a^b\left((x-b)^3+3(b-a)(x-b)^2+3(b-a)^2(x-b)+(b-a)^3\right)(x-b)^4dx\\&=\int_a^b\left((x-b)^7+3(b-a)(x-b)^6+3(b-a)^2(x-b)^5+(b-a)^3(x-b)^4\right)dx\\&=\left[\frac{(x-b)^8}{8}+\frac{3(b-a)(x-b)^7}{7}+\frac{3(b-a)^2(x-b)^6}{6}+\frac{(b-a)^3(x-b)^5}{5}\right]_a^b\\&=-\left(\frac{(a-b)^8}{8}+\frac{3(b-a)(a-b)^7}{7}+\frac{3(b-a)^2(a-b)^6}{6}+\frac{(b-a)^3(a-b)^5}{5}\right)\\&=-\left(\frac 18-\frac 37+\frac 36-\frac 15\right)(a-b)^8\\&=\frac{1}{280}(a-b)^8\end{align}$$
A: By integration by parts,
\begin{align}
\int_a^b(x-a)^3 (x-b)^4 dx &= \left[\frac{1}{5}(x-a)^3(x-b)^5\right]_a^b-\int_a^b \frac{3}{5}(x-a)^2(x-b)^5 dx\\
&=-\frac{3}{5}\left(\left[\frac{1}{6}(x-a)^2(x-b)^6\right]_a^b-\int_a^b \frac{1}{3}(x-a)(x-b)^6dx\right)\\
&=\frac{1}{5}\left(\left[\frac{1}{7}(x-a)(x-b)^7\right]_a^b-\int_a^b \frac{1}{7}(x-b)^7 dx\right)\\
&=-\frac{1}{35}\left[\frac{1}{8}(x-b)^8\right]_a^b\\
&=\frac{1}{280}(a-b)^8
\end{align}
A: By the linear change of variable that sends: $a \rightarrow 0, b \rightarrow 1$ (exercise!) you have a Beta integral https://en.wikipedia.org/wiki/Beta_function
whose value is $B(4,5)=\dfrac{3!4!}{8!}$ up to the factor (function of $a$ and $b$) that comes from the change of variable.
Edit: More precisely, the change of variable is 
$x=a+(b-a)t$ which gives for the differential element: $dx=(b-a)t$.
Thus the integral becomes: 
$$\int_0^1 \left((b-a)t\right)^3\left((b-a)(1-t)\right)^4 (b-a)dt$$
$$=(b-a)^8\int_0^1 t^3\left(1-t\right)^4 dt=(b-a)^8 B(4,5)=(b-a)^8 \dfrac{1}{288}$$
A: An alternative approach is to write $F(a,b) = \int^b_a(x-a)^3(b-x)^4 \,dx$ and then use
$\frac{\partial^7 F(a,b)}{\partial a^3 \partial b^4} \alpha \int^b_a \,dx = b-a.$
One point, since the limits also involve $b, a$ you would usually have to include a term that takes this into account, but in this case the integrand vanishes at these points. As well, you need to provide an argument that sets the arbitrary constants to zero for each integration over $a, b$ to get back $F(a,b)$.
