$$\int_0^{\pi/2} \sin^{2n - 1}x \, dx = \frac{2\cdot4\cdot6\cdots2n}{3\cdot5\cdot7\cdots(2n + 1)}$$ I have made the integration by parts substitutions as follows: $$u = \sin^{2n}(x) \text{ and } dv = \sin x\,dx$$
$$du = (2n)\sin(x)^{2n-1} \cos x \, dx$$
$$v = -\cos x$$
I am not sure if these are the correct substitutions to make and I am also unsure of the rest of the methods to complete the proof.
$$\int_0^{\pi/2} \sin^n x \, dx = \frac{n-1}{n} \int_0^{\pi/2} \sin^{n-2} x\,dx$$
$$u = \sin^{n-1} x \, dv = \sin x\,dx$$
$$du = (n-1)\sin^{n-2}(x)\cos x \, dx$$
$$v=-\cos x$$
I am pretty confident with these substitutions I just had a hard time with the rest of it.
\begin{align} \int\sin^{2n+1}x\,dx &=\int\sin^{2n}x\sin x\,dx \\[6px] &=-\cos x\sin^{2n}x+2n\int\cos^2x\sin^{2n-1}x\,dx \\[6px] &=-\cos x\sin^{2n}x+2n\int\sin^{2n-1}x\,dx-2n\int\sin^{2n+1}x\,dx \end{align} Therefore $$ (2n+1)\int\sin^{2n+1}x\,dx= -\cos x\sin^{2n}x+2n\int\sin^{2n-1}x\,dx $$ Hence $$ (2n+1)\int_0^{\pi/2}\sin^{2n+1}x\,dx= \Bigl[-\cos x\sin^{2n}x\Bigr]_0^{\pi/2}+2n\int_0^{\pi/2}\sin^{2n-1}x\,dx $$ and, finally, $$ \int_0^{\pi/2}\sin^{2n+1}x\,dx= \frac{2n}{2n+1}\int_0^{\pi/2}\sin^{2n-1}x\,dx $$ Now apply the induction hypothesis.