How Many Times Do I Need To Integrate?

Relevant wiki: Induction - Recurrence Relations

$\int_{0}^{\frac{\pi}{2}} \cos^n (\theta) d\theta=\int_{0}^{\frac{\pi}{2}} \sin^n (\theta) d\theta=I_n~(\text{Let})$

$I_n=\dfrac{n-1}{n}(I_{n-2})~~(**)$

Hence,

$\small{I_n=\left(\dfrac{n-1}{n}\right)\times\left(\dfrac{n-3}{n-2}\right)\times\left(\dfrac{n-5}{n-4}\right)\times\cdots\times\left(\dfrac{1}{2}\right)\times \underbrace{I_0}_{\large \pi/2}}$

$\small{=\dfrac{n!}{(n\times (n-2)\times (n-4)\cdots\times 4\times 2)^2}}\times \dfrac{\pi}2$

$\small{=\dfrac{n!}{2^n(\frac n2\times (\frac n2-1)\times (\frac n2-2)\cdots\times 2\times 1)^2}}\times \dfrac{\pi}2$

$\large =\dfrac{n!}{\left(\left(n/2\right)!\right)^2}\dfrac{\pi}{2^{n+1}}$ $\large =\dbinom{n}{n/2} \frac{\pi}{2^{n+1}}$

Consider $w=e^{\theta i}$ . Then $\cos^n \theta = \left(\dfrac{w+\frac{1}{w}}{2}\right)^n$ . If we expand it we got, considering that $n$ is even:

$\cos^n \theta = \dfrac{\dbinom{n}{0}(w^n+\frac{1}{w^n})+\dbinom{n}{1}(w^{n-2}+\frac{1}{w^{n-2}})+\dbinom{n}{2}(w^{n-4}+\frac{1}{w^{n-4}})+\cdots+\dbinom{n}{n/2}}{2^{n}}$

$\cos^n \theta = \dfrac{\dbinom{n}{0}\cos(n\theta)+\dbinom{n}{1}\cos((n-2)\theta)+\dbinom{n}{2}\cos((n-4)\theta)+\cdots+\dfrac{1}{2}\dbinom{n}{n/2}}{2^{n-1}}$

Then, the integral becomes:

$\dfrac{1}{2^{n-1}}\left[\dfrac{\binom{n}{0}}{n}\sin(n\theta)+\dfrac{\binom{n}{1}}{n-2}\sin((n-2)\theta)+\dfrac{\binom{n}{2}}{n-4}\sin((n-4)\theta)+\cdots+\dfrac{1}{2}\dbinom{n}{n/2}\theta\right]_{0}^{\frac{\pi}{2}}$

Note that $\sin\left(k\cdot\dfrac{\pi}{2}\right)=0$ for even $k$ , so the integral becomes simply $\boxed{\dbinom{n}{n/2}\dfrac{\pi}{2^{n+1}}}$ .

\begin{aligned} I & = \int_0^\frac \pi 2 \cos^\color{#3D99F6}{n} x \ dx \quad \quad \small \color{#3D99F6}{n \text{ is an even positive integer}} \\ & = \int_0^\frac \pi 2 \sin^0 x \cos^n x \ dx \\ & = \frac12 \color{#3D99F6}{B\left(\frac12, \frac {n + 1}2 \right) \quad \quad \small B(m,n) \text{ is beta function}} \\ & = \frac {\color{#3D99F6}{\Gamma \left(\frac12 \right)} \color{#D61F06}{\Gamma \left(\frac {n+ 1}2 \right)}}{2\Gamma \left(\frac n2 + 1 \right)} \quad \quad \small \color{#3D99F6}{\Gamma (x) \text{ is gamma function}} \\ & = \frac {\color{#3D99F6}{\sqrt{\pi}}\cdot \color{#D61F06}{(n-1)!!\sqrt{\pi}}}{2\cdot \color{#D61F06}{2^\frac n2} \cdot \left(\frac n2 \right)!} \\ & = \frac {n! \pi}{2 \cdot 2^\frac n2 \cdot 2^\frac n2 \cdot \left(\frac n2 \right)!\cdot \left(\frac n2 \right)!} \\ & = \boxed{\displaystyle \binom {n}{n/2} \frac{\pi}{2^{n+1}}} \end{aligned}

Wallis' integrals !!!!!