The integral
$$\int_0^2 x^5 (2-x)^4 \text dx$$
Can be integrated using beta function by substituting $x=2\sin^{2}\theta$ and using the definition of the beta function
$$\beta(x,y)=2\int_{0}^{\frac{\pi}{2}}\sin^{2x-1}\theta\cos^{2y-1}\theta\, \text d\theta$$
Now suppose that our bounds have changed to $0$ and $1$. Is there a substitution that will make this new integral solvable by means of the beta function?
$$\int_0^1x^5(2-x)^4\text dx$$
With the listed integral, substitution and definition in hand, we have $x = 2\sin^2\theta,\text dx = 4\sin\theta\cos\theta\text { d}\theta$ from which we get
$$\int_0^2 x^5 (2-x)^4 \text dx = 2^{11}\int_{0}^{\frac{\pi}{2}}\sin^{11}\theta\cos^{9}\theta\, \text d\theta = 2^{10}\beta(6,5)$$
If the integral has a bounds-shift, and we wish to know what happens if the upper bound only goes to $1$, then (due to lack of symmetry of the function across the interval) we need to calculate
$$\int_1^2 x^5(2-x)^4\text dx = \int_0^1(1+y)^5(1-y)^4\text dy\\ =\int_0^1(1+y)(1-y^2)^4\text dy = \int_0^1(1-y^2)^4\text dy+\int_0^1y(1-y^2)^4\text dy\\ =\int_0^1(1-y^2)^4\text dy + \frac 12\beta(1,5)$$
Now, with $I = \int_0^1(1-y^2)^4\text dy = \int_1^2x^4(2-x)^4\text dx$ left, we have:
$$2I = \int_0^2x^4(2-x)^4\text dx = 2^9\beta(5,5)$$
Adding everything together, we get
$$\int_0^1x^5(2-x)^4\text dx = 2^{10}\beta(6,5)-2^8\beta(5,5)-\frac 12\beta(1,5)$$
If we manipulate this further using the factorial form, we find that
$$2^{10}\beta(6,5)-2^8\beta(5,5)-\frac 12\beta(1,5)\\ =2^{10}{5!4!\over 10!}-2^8{4!4!\over 9!}-{4!\over 2\cdot 5!}\\ =2^9{5!4!\over 10!}-2*7*9{5!4!\over 10!}=2\cdot 193\beta(6,5)$$
So there must exist a transformation such that
$$\int_0^1x^5(2-x)^4\text dx = 386\int_0^1u^5(1-u)^4\text du$$
This does not mean that such a transformation will be easy or obvious. Since $193=256-63=128+64+1$ is prime, it has only the above summation relationship to the original.