$\int_{-\pi/2}^{\pi/2} dx \, \sin^{2n} x $

Question

While reading a physics paper I came cross the following set of integrals.

$$\int_{-\pi/2}^{\pi/2} dx \, \sin^{2n} x $$

I tried using De Moivre identity but not sure about the conclusion:

$$ \int_{-\pi/2}^{\pi/2} dx \, \left(\frac{e^{ix} - e^{-ix}}{2}\right)^{2n} = \frac{1}{2^n} \int_{-\pi/2}^{\pi/2} dx \, \left(e^{2ix} + e^{-2ix} - 2 \right)^{n} $$

The idea is into integrate over the entire period take the middle term of the binomial expansion

\begin{eqnarray} \frac{1}{2^n} \int_{-\pi}^{\pi} dx \, \left(e^{ix} + e^{-ix} - 2 \right)^{n} &=& \frac{1}{2^n} \int_{-\pi}^{\pi} \sum_{k=0}^n dx \, (e^{ix} + e^{-ix})^k 2^{n-k} \binom{n}{k} \\ &=& \sum_{k=0}^n 2^{-k}\binom{2k}{k} \binom{n}{k} \end{eqnarray}

I couldn't find the correct page in the table of definite integrals. Can anyone help with the derivation and the correct formula?

$$ \int_0^{\pi/2} \sin^{2n} dx = \frac{1}{2^{2n}}\binom{2m}{m} \frac{\pi}{2} $$

Answer 1

Since $$\int_{-\pi/2}^{\pi/2} e^{2ni\theta}\,d\theta = 0 $$ for any $n\in\mathbb{Z}\setminus\{0\}$, we have, by expanding $(e^{ix}-e^{-ix})^{2n}$ with the binomial theorem, $$I=\int_{-\pi/2}^{\pi/2}\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{2n}\,dx = \frac{(-1)^n}{4^n}\binom{2n}{n}(-1)^n\frac{\pi}{2}=\frac{\pi}{2\cdot 4^n}\binom{2n}{n}.$$

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If someone is looking for an overkill, by using the Euler Beta function we have: $$ I = 2\int_{0}^{1} t^{2n}(1-t^2)^{-1/2}\,dt = \int_{0}^{1} u^{n-1/2}(1-u)^{-1/2}\,du = \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+1)}.$$

Answer 2

This is known as Wallis' integral. The proof is based on an integration by parts and then a recursive formula.

Answer 3

Another approach$$I_{2n}=∫_{-\frac{π}{2}}^{\frac{π}{2}}\sin ^{2n}xdx=2∫_{0}^{\frac{π}{2}}\sin ^{2n}xdx=\\2∫_{0}^{\frac{π}{2}}\sin ^{2n-1}x\sin xdx=-2\cos x \sin x |_{0}^{π/2}-2∫_{0}^{\frac{π}{2}}\left(2n-1\right)\sin ^{2n-2}x\cos xdx\\=0+2∫_{0}^{\frac{π}{2}}\left(2n-1\right)\sin ^{2n-2}x\cos ^{2}xdx=\\2∫_{0}^{\frac{π}{2}}\left(2n-1\right)\sin ^{2n-2}x\left(1-\sin ^{2}x\right)dx=\\2\left(2n-1\right)∫_{0}^{\frac{π}{2}}\left(\sin ^{2n-2}x-\sin ^{2n}x\right)dx=\\2\left(2n-1\right)\left(I_{2n-2}-I_{2n}\right)$$now we have a recursive relation $$I_{2n}=2\left(2n-1\right)\left(I_{2n-2}-I_{2n}\right)\\\left(4n-2+1\right)I_{2n}=\left(4n-2\right)I_{2n-2}\\I_{2n}=\frac{\left(4n-2\right)}{\left(4n-1\right)}I_{2n-2}\\I_{0}=π\\$$