<s>[INST] 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)?
 [/INST] How about the following way? (though I'm not sure if you like it)
$$\begin{align}\\&\int_{a}^{b}(\color{red}{x-a})^3(b-x)^4dx\\&=\int_{a}^{b}(\color{red}{x-b+b-a})^3(x-b)^4dx\\&=\int_a^b\left((x-b)^3+3(b-a)(x-b)^2+3(b-a)^2(x-b)+(b-a)^3\right)(x-b)^4dx\\&=\int_a^b\left((x-b)^7+3(b-a)(x-b)^6+3(b-a)^2(x-b)^5+(b-a)^3(x-b)^4\right)dx\\&=\left[\frac{(x-b)^8}{8}+\frac{3(b-a)(x-b)^7}{7}+\frac{3(b-a)^2(x-b)^6}{6}+\frac{(b-a)^3(x-b)^5}{5}\right]_a^b\\&=-\left(\frac{(a-b)^8}{8}+\frac{3(b-a)(a-b)^7}{7}+\frac{3(b-a)^2(a-b)^6}{6}+\frac{(b-a)^3(a-b)^5}{5}\right)\\&=-\left(\frac 18-\frac 37+\frac 36-\frac 15\right)(a-b)^8\\&=\frac{1}{280}(a-b)^8\end{align}$$
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