<s>[INST] Integral of $\int_0^{\pi/2} \ (\sin x)^7\ (\cos x)^5 \mathrm{d} x$ I am trying to find this by using integration by parts but I am not sure how to do it.
$$\int_0^{\pi/2}  (\sin x)^7 (\cos x)^5 \mathrm{d} x$$
I tried rewriting as 
$$\int_0^{\pi/2} \sin x\cdot\ (\sin x)^6\cdot\ (\cos x)^5 \mathrm{d} x = \int_0^{\pi/2}\sin x(1-\ (\cos x)^3)\cdot\ (\cos x)^5 \mathrm{d} x$$
but that seems to only give me a very, very long loop that doesn't help me at all. How do I proceed?
$$\int_0^{\pi/2} \sin x\cdot (\sin x)^6\cdot (\cos x)^5 \mathrm{d} x = \sin x(1- (\cos x)^3)\cdot (\cos x)^5 \mathrm{d} x$$
$u = \cos x$, then $du = -\sin xdx$
$\int \frac{-u^6}{6} \mathrm{d} u - \int \frac{-u^9}{9} \mathrm{d} u$
From here it looks like I have an incredibly long string of $u$ substitutions to make to get to something I can find an antiderivative for.
 [/INST] Beta version:
$$\begin{aligned}\int_{0}^{\frac{\pi}{2}}\left(\sin x\right)^{7}\left(\cos x\right)^{5}dx & =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(\sin x\right)^{6}\left(\cos x\right)^{4}2\sin x\cos xdx\\
 & =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(\left(\sin x\right)^{2}\right)^{3}\left(1-\left(\sin x\right)^{2}\right)^{2}d\left(\left(\sin x\right)^{2}\right)\\
 & =\frac{1}{2}\int_{0}^{1}t^{3}\left(1-t\right)^{2}dt=\frac{1}{2}\int_{0}^{1}t^{4-1}\left(1-t\right)^{3-1}dt\\
 & =\frac{1}{2}B\left(4,3\right)=\frac{1}{2}\frac{\Gamma\left(4\right)\Gamma\left(3\right)}{\Gamma\left(7\right)}=\frac{1}{2}\frac{6\cdot2}{720}=\frac{1}{120}
\end{aligned}$$
</s>