How may one prove something similar as in here but from $0$ to $\pi$ and without using the Residue Theorem?
I was told to consider the contour integral $$\int_{|z|=1}(z-\frac{1}{z})^{2n}dz/z $$
and so I parametrise it with $z=e^{it} $ and got $$\int_0^\pi\sin^{2n}t dt=\frac{1}{4^{n}i^{2n+1}}\int_{|z|=1}(z-\frac{1}{z})^{2n}dz/z$$ I am not sure what I did is helpful, even if so I have no idea how to continue from there.
One idea would be to use the binomial theorem for $(z-1/z)^{2n}$, and then for each $ -n \leq k \leq n$ calculate, if you don't know it yet $$\int_{|z|=1} z^kdz$$
You can also calculate $\int_0^\pi\sin^{2n}t dt$ without complex Analysis. Use integration by parts with $f =\sin^{n-1}(t)$ and $g'=\sin(t)$, and then use $\cos^2(t)=1-\sin^2(t)$. This gives you a relation between $\int_0^\pi\sin^{2n}t dt$ and $\int_0^\pi\sin^{2n-2}t dt$, and then by induction you can prove the formula you seek.
Added
$$(z-1/z)^{2n}=\sum_{k=0}^{2n} \binom{2n}{k} z^{2n-k}(-1)^k z^{-k}=\sum_{k=0}^{2n} \binom{2n}{k} z^{2n-2k}(-1)^k $$
Therefore $$\sum_{k=0}^{2n} \binom{2n}{k} z^{2n-k}(-1)^k z^{-k}=\sum_{k=0}^{2n} \binom{2n}{k}(-1)^k \int_{|z|=1} z^{2n-2k-1} dz$$
Now, just calculate $\int_{|z|=1} z^{2n-2k-1}$ you will see that this is zero for all $k \neq n$.
