A recursive formula for $a_n$ = $\int_0^{\pi/2} \sin^{2n}(x)dx$, namely $a_n = \frac{2n-1}{2n} a_{n-1}$ Where does the $\frac{2n-1}{2n}$ come from? 
I've tried using integration by parts and got $\int \sin^{2n}(x)dx = \frac {\cos^3 x}{3} +\cos x +C$, which doesn't have any connection with $\frac{2n-1}{2n}$.
Here's my derivation of $\int \sin^{2n}(x)dx = \frac {\cos^3 x}{3} +\cos x +C$:
$\sin^{2n+1}xdx=\int(1-\cos^2x)\sin xdx=\int -(1-u^2)du=\int(u^2-1)du=\frac{u^3}{3}+u+C=\frac{\cos^3x}{3}+\cos x +C$
where 
$u=\cos x$;$du=-\sin x dx$
Credits to Xiang, Z. for the question.
 A: Given the identity $$\int \sin^n x dx = - \frac{\sin^{n-1} x \cos x}{n}+\frac{n-1}{n}\int \sin^{n-2} xdx$$ plugging in $2n$ yields $$\int \sin^{2n} x dx = - \frac{\sin^{2n-1} x \cos x}{2n}+\frac{2n-1}{2n}\int \sin^{2n-2} xdx$$
Since
$$\int_0^{\pi/2} \sin^{2n} x dx = - \frac{\sin^{2n-1} x \cos x}{2n}|_0^{\pi/2}+\frac{2n-1}{2n}\int_0^{\pi/2} \sin^{2n-2} xdx$$ and $\frac{\sin^{2n-1} x \cos x}{2n}|_0^{\pi/2}=0$ for $n \ge 1$, we get $$\int_0^{\pi/2} \sin^{2n} x dx = \frac{2n-1}{2n}\int_0^{\pi/2} \sin^{2n-2} xdx$$
(We only care about $n \ge 1$ because in the original question, $a_0=\frac{\pi}{2}$ is given and only integer values of n with $n \ge 1$ need to satisfy $a_n=\frac{2n-1}{2n}a_{n-1}$.)
A: HINT:
Using this, $$\int_0^{\frac\pi2}\sin^{2n}xdx$$
$$=\int_0^{\frac\pi2}\sin^{2n}\left(\frac\pi2+0-x\right)dx$$ as $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$
$$=\int_0^{\frac\pi2}\cos^{2n}xdx$$
$$=\frac{\cos^{n-1}x\sin x}n\big|_0^{\frac\pi2}+\frac{2n-1}{2n}\int_0^{\frac\pi2}\cos^{2n-2}xdx$$
$$=0+\frac{2n-1}{2n} \int_0^{\frac\pi2}\cos^{2n-2}\left(\frac\pi2+0-x\right)dx\text{ for } n-1>0\iff n>1$$
$$=\frac{2n-1}{2n} \int_0^{\frac\pi2}\sin^{2n-2}xdx$$
A: Consider the integral
$$I(a,b)=\int_0^{\pi/2}\sin^ax\ \cos^bx\ dx$$
$t=\sin^2x$:
$\therefore dt=2\sin x\cos x\ dx\\\therefore dx=\frac12t^{-1/2}(1-t)^{-1/2}dt\\\therefore x=0\mapsto t=0\\\therefore x=\pi/2\mapsto t=1$
$$I(a,b)=\int_0^1 t^{a/2}(1-t)^{b/2}\frac12t^{-1/2}(1-t)^{-1/2}dt$$
$$I(a,b)=\frac12\int_0^1 t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}dt$$
$$I(a,b)=\frac12\int_0^1 t^{\frac{a+1}2-1}(1-t)^{\frac{b+1}2-1}dt$$
Note the definition of the Beta function:
$$B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=B(b,a)$$
Thus
$$I(a,b)=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}=I(b,a)$$
Plugging in:
$$I(2n,0)=\frac{\Gamma(n+\frac12)\Gamma(\frac12)}{2\Gamma(n+1)}$$
$$I(2n,0)=\frac{(n-\frac12)\Gamma(n-\frac12)\Gamma(\frac12)}{2n\Gamma(n)}$$
$$I(2n,0)=\frac{2n-1}{2n}\frac{\Gamma(n-\frac12)\Gamma(\frac12)}{2\Gamma(n)}$$
$$I(2n,0)=\frac{2n-1}{2n}I(2n-1,0)$$
QED
