Gamma and Beta Functions

Question

I have this example
$$ \int_{0}^{\pi/2}\sin^{5}x\cos^{7}x dx $$ and i know formula $$ \frac{1}{2}B\left(\frac{m+1}{2} , \frac{n+1}{2}\right) = \frac{1}{2}B\left(\frac{5+1}{2} , \frac{7+1}{2}\right) = \frac{1}{2}B(3,4) = 6 $$ what next step i must to do? Thanks

Answer 1

Using the form of the Beta function as \begin{align} B(x,y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} dt \end{align} then the substitution $t = \cos^{2}\theta$ leads to the integral \begin{align} B(x,y) = 2 \int_{0}^{\pi/2} \sin^{2x-1}\theta \, \cos^{2y-1}\theta \, d\theta. \end{align} Now using the definition of $B(x,y)$ in terms of the Gamma function, namely, \begin{align} B(x,y) = \frac{ \Gamma(x) \Gamma(y) }{ \Gamma(x+y) } \end{align} then the desired integral can be seen to be \begin{align} \int_{0}^{\pi/2} \sin^{5}\theta \, \cos^{7}\theta \, d\theta &= \frac{1}{2} \, B(3, 4) = \frac{\Gamma(3) \Gamma(4) }{ 2 \Gamma(3+4) } = \frac{ 2! \, 3!}{2 \cdot 6!} = \frac{1}{4 \cdot 5 \cdot 6} = \frac{1}{120}. \end{align}