I'm trying to show the following.
$$ \int^1_0 x^3 \sqrt{x} \sqrt{1-x} dx = \frac{\pi}{5!} \frac{1\cdot3\cdot5\cdot7}{2^5} $$
This is a problem regarding contour integration. My complex analysis knowledge is mostly self taught, so I don't have all the fine details sorted out. From my research on the internet, I think the approach is to construct $f(z)$ so that you it has a branch cut on $[0,1]$ and use branches of the logarithm function. Would appreciate any help or guidance.
You can solve this without complex analysis, using a trigonometric substitution. Let $x=\sin ^2 \theta $, then $dx = 2 \sin \theta \cos \theta d\theta$ and you have $$ \int_0^1 x^3 \sqrt{x} \sqrt{1-x} dx = \int_0^{\pi/2} 2 \sin^8 \theta \cos^2 \theta d\theta = 2 \int_0^{\pi/2} \sin^8 \theta d\theta -2\int_0^{\pi/2} \sin^{10} \theta d\theta $$ If you let $$ I_n = \int_0^{\pi/2} sin^n \theta d\theta $$ your answer becomes $2I_8 - 2I_{10}$.
$I_n$ is a well-known general integral. You can solve it by finding a reduction formula by integrating by parts twice and evaluating it directly for $n=0$ and $n=1$.
That should give you the answer. If you need any more details I omitted near the end, tell me.