What is $\int_0^{\pi/2}\sin^7(\theta)\cos^5(\theta)d\theta$

I think you complicated the last part, after all you are integrating a polynomial.

$\displaystyle \int_0^1 u^3(1-u)^2\mathop{du}=\int_0^1 (u^3-2u^4+u^5)\mathop{du}=\left[\frac{u^4}4-2\frac{u^5}5+\frac{u^6}6\right]_0^1=\frac 14-\frac 25+\frac 16=\frac 1{60}$

Also you dropped the coeff $\dfrac 12$ from $\dfrac{du}2$, the result should be $\dfrac 1{120}$

Note that: $$B(m+1,n+1)=2\int_0^{\pi/2}\cos^{2m+1}(\theta)\sin^{2n+1}(\theta)d\theta=\frac{m!n!}{(m+n+1)!}$$

I would just do $u=\sin\theta$ and $\mathrm du=\cos\theta\,\mathrm d\theta$. So\begin{align}\int_0^{\frac\pi2}\sin^7(\theta)\cos^5(\theta)\,\mathrm d\theta&=\int_0^{\frac\pi2}\sin^7(\theta)\bigl(1-\sin^2(\theta)\bigr)^2\cos(\theta)\,\mathrm d\theta\\&=\int_0^1u^7(1-u^2)^2\,\mathrm du.\end{align}I think that it's simpler.