Beta Distribution and how it is related to this question 
Let $f(x) = k(\sin x)^5(1-\sin x)^7$ if $0 \lt x \lt \pi/2$ and $0$ otherwise. Find the value of $k$ that makes $f(x)$ a density function. 

I'm struggling to understand how this relates to the Beta distribution and how to proceed with this question.
 A: The integral of $f$ can be expressed as a Beta function times a hypergeometric function.  This suggests $f$ is not the density of any particular Beta distribution, but that it is indeed related.
To evaluate $k$ it's simpler and more elementary to use the substitution $y = \sin^2(x),$ $\mathrm{d}y = 2\sin(x)\cos(x)\mathrm{d}x$ and observe $\cos(x) = (1-\sin^2(x))^{1/2}$ to evaluate integrals of the form
$$\int_0^{\pi/2} \sin^m(x)\cos^n(x)\mathrm{d}x = \frac{1}{2}\int_0^1 y^{(m-1)/2} (1-y)^{(n-1)/2}\mathrm{d}y = \frac{1}{2}B\left(\frac{m+1}{2}, \frac{n+1}{2}\right)$$
and then apply the Binomial Theorem to expand $(1-\sin(x))^7,$ producing
$$\eqalign{
\int_0^{\pi/2} f(x)\mathrm{d}x &= k \int_0^{\pi/2} \sin^5(x)(1-\sin(x))^7 \mathrm{d}x \\
&=k \sum_{j=0}^7 (-1)^j \binom{7}{j}\int_0^{\pi/2}\sin^{5+j}(x)\mathrm{d}x \\
&= \frac{k}{2} \sum_{j=0}^7 (-1)^j \binom{7}{j} B\left(\frac{5+j+1}{2}, \frac{1}{2}\right).
}$$
The alternating binomial coefficients $\binom{7}{j}$ in this sum are $$\binom{7}{j} = (1,-7,21,-35,35,-21,7,-1)$$ while the Beta function values are $$B\left(\frac{5+j+1}{2}, \frac{1}{2}\right) = \left(\frac{16}{15},\frac{5 \pi }{16},\frac{32}{35},\frac{35 \pi }{128},\frac{256}{315},\frac{63 \pi }{256},\frac{512}{693},\frac{231 \pi }{1024}\right),$$
giving
$$1 = \int_0^{\pi/2} f(x)\mathrm{d}x = k \color{blue}{\frac{1}{2}\left(\frac{26672}{495} -\frac{17563}{1024}\pi\right)}.$$
Thus, $k$ is the reciprocal of the blue quantity.
