$\sin^{2n}x$ can be written as $(1 – \cos^2x)^n$. Let
$$A_{a,n}=\int\limits_{0}^{\pi} (a – \cos^{2}(x))^{n}\,\mathrm dx$$
Which for when $a = 1$ gives the integral in question and
$$B_{a,n} = \int\limits_{0}^{\pi} (a – \sin^{2}(x))^{n}\,\mathrm dx$$
To use Feynman trick for $A_{n}$, setting $a = 0$ for the constant of integration gives $B_{0,n}$. Then differentiating under the integral with respect to $a$ gives
$$\frac {\partial f}{\partial a}=\int\limits_0^{\pi}(a-\cos^2x)^{n-1}\,\mathrm dx\qquad\implies\qquad\frac {\partial f}{\partial a}=A_{a,n-1}$$
You could get the same result when you switch $A_{a,n}$ and $B_{a,n}$. $A_{0,0}$ and $B_{0,0}$ would simply be both $\pi$.
With the above in mind, is there anyway to use Feynman trick, reduction formula, and sequences to solve
$$\int\limits_{0}^{\pi}\sin^{2n}x\,\mathrm dx$$
If there is another way, please don't tell me, I want some more time to figure out this problem. Thank you
Best Answer
I don't know if this is what you want. Let $$ I_n=\int_0^\pi\sin^{2n}(x)dx. $$ Then by integration by parts, one has \begin{eqnarray} I_{n+1}&=&-\int_0^\pi\sin^{2n+1}(x)d\cos (x)\\ &=&(2n+1)\int_0^\pi\cos^2(x)\sin^{2n}(x)dx\\ &=&(2n+1)\int_0^\pi\sin^{2n}(x)dx-(2n+1)\int_0^\pi\sin^{2n+2}(x)dx\\ &=&(2n+1)I_{n+1}-(2n+1)I_n \end{eqnarray} and hence $$ I_{n+1}=\frac{2n+1}{2n}I_n. $$ So $$ I_n=\frac{2n-1}{2(n-1)}\cdot\frac{2n-3}{2(n-2)}\cdots\frac{3}{2}I_1=\frac{(2n-1)!!}{(2n-2)!!}\frac{\pi}{2}. $$