$I=\int_3^5(x-3)^3(5-x)^5 d x$
Let, $x=3 \cos ^2 \theta+5 \sin ^2 \theta$, so at $x=3, \theta=0$
and at $x=5, \theta=\frac{\pi}{2}$ and $d x=2 \sin 2 \theta d \theta$
Therefore,
$I=\int_0^{\pi / 2}\left(2 \sin ^2 \theta\right)^3\left(2 \cos ^2 \theta\right)^5 2(\sin 2 \theta) d \theta$
$=2^{10} \int_0^{\pi / 2} \sin ^7 \theta \cos ^{11} \theta d \theta$
$=\frac{2^{10} \times(7-1)(7-3)(7-5) \times(11-1)(11-3)(11-5)(11-7)(11-9)}{18 \times 16 \times 14 \times 12 \times 10 \times 8 \times 6 \times 4 \times 2}$
$=2^{10} \frac{(6 \times 4 \times 2)(10 \times 8 \times 6 \times 4 \times 2)}{18 \times 16 \times 14 \times 12 \times 10 \times 8 \times 6 \times 4 \times 2}$
$=2^{10} \frac{6 \times 4 \times 2}{18 \times 16 \times 14 \times 12}=2^{10} \frac{1}{16 \times 7 \times 9}=\frac{64}{63}$