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)?

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}
