1
$\begingroup$

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 $$I_{2n}=\frac{2n-1}{2n}I_{2n-2}$$

$\endgroup$
1
$\begingroup$

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 $I_{2n} = \frac{2n-1}{2n} I_{2n-2}$.

$\endgroup$
  • $\begingroup$ can't we prove it using the Green's theorem? thanks for your answer btw $\endgroup$ – user159234 Jul 13 '17 at 12:16
  • $\begingroup$ I fixed my answer. could you take a look at it and tell me if I'm right? I used your idea. $\endgroup$ – user159234 Jul 13 '17 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.