Step 1: Recognise the Beta function shape.
Any integral that looks like $\int_0^1 x^{m-1}(1-x)^{n-1}\,dx$ is the Beta function $\beta(m,n)$. Our integral $\int_0^1 x^{5/2}(1-x)^{3/2}\,dx$ matches this pattern perfectly, so we just need to read off $m$ and $n$.
Step 2: Read off $m$ and $n$.
Matching powers: $m - 1 = \frac{5}{2}$ gives $m = \frac{7}{2}$, and $n - 1 = \frac{3}{2}$ gives $n = \frac{5}{2}$.
Step 3: Write it with Gamma functions.
The Beta to Gamma rule says $\beta(m,n) = \dfrac{\Gamma(m)\,\Gamma(n)}{\Gamma(m+n)}$. Here $m+n = \frac{7}{2}+\frac{5}{2}=6$, so \[ I = \frac{\Gamma\!\left(\tfrac{7}{2}\right)\Gamma\!\left(\tfrac{5}{2}\right)}{\Gamma(6)} \] and $\Gamma(6) = 5! = 120$.
Step 4: Evaluate the half-integer Gammas.
Use $\Gamma(p+1)=p\,\Gamma(p)$ and $\Gamma(\tfrac{1}{2})=\sqrt{\pi}$. So $\Gamma(\tfrac{7}{2}) = \frac{5}{2}\cdot\frac{3}{2}\cdot\frac{1}{2}\sqrt{\pi} = \frac{15}{8}\sqrt{\pi}$ and $\Gamma(\tfrac{5}{2}) = \frac{3}{2}\cdot\frac{1}{2}\sqrt{\pi} = \frac{3}{4}\sqrt{\pi}$.
Step 5: Multiply the numerator.
$\Gamma(\tfrac{7}{2})\,\Gamma(\tfrac{5}{2}) = \frac{15}{8}\sqrt{\pi}\cdot\frac{3}{4}\sqrt{\pi} = \frac{45}{32}\pi$, since $\sqrt{\pi}\cdot\sqrt{\pi}=\pi$.
Step 6: Divide by $120$ and simplify.
\[ I = \frac{\frac{45}{32}\pi}{120} = \frac{45\pi}{3840} = \frac{3\pi}{256} \]
\[ \boxed{\dfrac{3\pi}{256}} \]