A recursive formula for $a_n$ = $\int_0^{\pi/2} \sin^{2n}(x)dx$, namely $a_n = \frac{2n-1}{2n} a_{n-1}$

Question

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.

Answer 1

HINT: Using this, $$\int_0^{\frac\pi2}\sin^{2n}xdx$$

$$=\int_0^{\frac\pi2}\sin^{2n}\left(\frac\pi2+0-x\right)dx$$ as $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$

$$=\int_0^{\frac\pi2}\cos^{2n}xdx$$

$$=\frac{\cos^{n-1}x\sin x}n\big|_0^{\frac\pi2}+\frac{2n-1}{2n}\int_0^{\frac\pi2}\cos^{2n-2}xdx$$

$$=0+\frac{2n-1}{2n} \int_0^{\frac\pi2}\cos^{2n-2}\left(\frac\pi2+0-x\right)dx\text{ for } n-1>0\iff n>1$$

$$=\frac{2n-1}{2n} \int_0^{\frac\pi2}\sin^{2n-2}xdx$$

Answer 2

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 $a_n=\frac{2n-1}{2n}a_{n-1}$.)