Integration by parts, proving inductive case ${1\over2}\int_{-\pi/2}^{\pi/2}cos^{2n-1}(x) dx$
Inductive step: Show that the $integral={(2n-2)(2n-4)...\over (2n-1)(2n-3)...}$ for $n\ge2$
$T(n+1)$=...
Attempted int. by parts using $u=cos^{2n}(x)dx$, $dv=cos(x)$,
but ended up with -2n/3 as the result. $uv$ integration goes to 0. Could someone point me in the right direction.

Note that $\frac{1}{2}\int_{-\pi/2}^{\pi/2}{\cos^{2n-1}(x)}dx=\frac{(2n-2)(2n-4)...4\times 2}{(2n-1)(2n-3)...5\times 3}$. 
Consider $\frac{1}{2}\int_{-\pi/2}^{\pi/2}{\cos^{2n+1}(x)}dx$. When using integration by part, let $u=\cos^{2n-1}(x)$ and $dv= \cos^2(x)dx$
