Prove $\int_0^{2\pi} \sin^{2n}t dt=\frac{2n-1}{2n}\int_0^{2\pi}\sin^{2n-2}tdt$ Prove $$\int_0^{2\pi} \sin^{2n}t dt=\frac{2n-1}{2n}\int_0^{2\pi}\sin^{2n-2}tdt$$ for all integer $n>0$.
My attempt: Let $x=\cos t$, $y=\sin t$. Then $\sin^{2n}tdt=y^{2n}(-\frac{dx}{y})$ since $dx=-\sin tdt=-ydt.$ So we have $$\int_0^{2\pi}\sin^{2n}tdt=\int_C-y^{2n-1}dx+0\cdot dy$$where $C$ is a unit circle $\{(x,y):x^2+y^2=1\}$. Now apply the Green's theorem: $$\int_C-y^{2n-1}dx+0\cdot dy=\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}(\frac{\partial}{\partial x}0-\frac{\partial}{\partial y}(-y^{2n-1}))dxdy\\=\int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}(2n-1)y^{2n-2}dxdy\\=(2n-1)\int_{-1}^1\int_{-\cos t}^{\cos t}\sin^{2n-2}t(-\sin t dt)(\cos tdt)$$ So, here, I have two $dt$'s, I think I did something wrong, but I don't know how to fix it. Any ideas?
EDIT: I'm supposed to use the Green's theorem.
EDIT2: since $$(2n-1)\int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}y^{2n-2}dxdy=(2n-1)\int_{-1}^1y^{2n-2}\cdot 2\sqrt{1-y^2}dy\\=2(2n-1)\int_0^{2\pi}\sin^{2n-2}t\cos^2tdt\\=2(2n-1)(I_{2n-2}-I_{2n})=I_{2n}$$ where $I_{2n}=\int_0^{2\pi}\sin^{2n-2}tdt$, so I have $$I_{2n}=\frac{4n-2}{4n-1}I_{2n-2}$$ but I'm suppose to have $$I_{2n}=\frac{2n-1}{2n}I_{2n-2}$$. I guess the $2$ shouldn't have come out during the computation, but can't make it gone. 
EDIT3: so, since we integrate $y$ from $-1$ to $1$, and $y=\sin t$, we are integrating $t$ from $-\pi$ to $\pi$. so we have $$(2n-1)\int_{-1}^1y^{2n-2}\cdot 2\sqrt{1-y^2}dy=2(2n-1)\int_{-\pi}^{\pi}\sin^{2n-2}t\cos^2tdt\\=2(2n-1)\cdot \frac12(I_{2n-2}-I_{2n})$$ hence $$(2n-1)(I_{2n-2}-I_{2n})=I_{2n}$$ and $$I_{2n}=\frac{2n-1}{2n}I_{2n-2}$$

Write $\int_0^{2\pi} \sin^{2n}x \, \mathrm{d}x = \int_0^{2\pi} \sin^{2n-1} x \sin x \, \mathrm{d}x$ and apply IBP with $u = \sin^{2n-1} x$ and $v = -\cos x$ so that $$\begin{align*}I_{2n} &= \big[(2n-1) \sin^{2n-2} x\cos x\big]_{2\pi}^0 + (2n-1)\int_0^{2\pi} \sin^{2n-2} x \cos^2 x \, \mathrm{d}x \\  & = (2n-1)\int_0^{2\pi} \sin^{2n-2} x (1-\sin^2 x) \, \mathrm{d}x \\ & = (2n-1)I_{2n-2} - (2n-1) I_{2n}\end{align*}$$
Hence $I_{2n} = \frac{2n-1}{2n} I_{2n-2}$.
