How do you go about proving that
integral of sin(X)^(2n+1) dx evaluated on the interval [0,pi/2]= pi/2 x [(2 x 4 x 6.......x (2n-1)) / (3 x 5 x 7...........x 2n)] ?
Using integration by parts to:
$\displaystyle I_{2n+1}=\displaystyle\int_{0}^{\pi/2}\sin ^{2n+1} x\;dx,\quad (u=\sin^{2n} x,\;dv=\sin x dx)$
you'll obtain the recurrence relation:
$\displaystyle I_{2n+1}=\dfrac{2n}{2n+1}I_{2n-1}$
Fernando Revilla