$\int_0^\pi{\cos^{2n}(x)dx,~ n \in \mathbb N}$ in easier way. I wanted to evaluate following integral:
$$\int_0^\pi{\cos^{2n}(x)dx,~ n \in \mathbb N}$$
I managed to do this by turning cosine into its complex form $\frac{e^{ix}+e^{-ix}}{2}$ and then using binomial theorem, and I got the following result:
$$\int_0^\pi{\cos^{2n}(x)dx=\frac{1}{2^{2n}}\frac{(2n)!}{(n!)^2}\pi,~ n \in \mathbb N}$$
However I wonder, whether I could achieve it without using complex numbers, but rather with some trigonometric identities, real analysis etc.
 A: \begin{align*}
I_{2n}&=\displaystyle\int_0^{\pi}\cos^{2n}(x)dx=\displaystyle\int_{0}^{\pi}\cos(x)\cos^{2n-1}(x)dx\\
&=\left[\sin(x)\cos^{2n-1}(x)\right]_0^{\pi}-\displaystyle\int_0^{\pi}\sin(x)(2n-1)(-\sin(x))\cos^{2n-2}(x)dx\\
&=(2n-1)I_{2n-1}-(2n-1)I_{2n}\\
&\implies I_{2n}=\frac{2n-1}{2n}I_{2n-2}\\
&\implies I_{2n}=\frac{(2n-1)!!}{(2n)!!}\pi\\
\end{align*}
and we have :
$$(2n)!!=2^n(n!)$$
and $$(2n+1)!!=\frac{(2n+2)!}{2^{n+1}(n+1)!}$$
Then you can find the one you want
A: Let us do integration by parts taking $\cos x$ as second function:
$$I_n=\int_{0}^{\pi} \cos^{2n-1} x ~ \cos x~dx=\cos^{2n-1} x \sin x|_{0}^{\pi}+(2n-1) \int \cos ^{2n-2} x \sin^2 x dx. $$
Use $\sin^ x=1-\cos^2 x$, then
$$\implies I_n=(2n-1)\int [\cos^{2n-2} x- \cos^{2n} x] sx$$
$$\implies (1+2n-1)I_n=(2n-1)I_{n-1} \implies I_n=\frac{(2n-1)}{2n}I_{n-1},$$
where $I_0=\pi$I_{n-1}
$$\implies I_1=\frac{1}{2}I_0, ~I_2=\frac{3}{4}I_1,~I_3=\frac{5}{6},~I_4=\frac{7}{8}I_3,....I_{n-1}=\frac{2n-3}{2n-2}I_{n-2},~I_n=\frac{2n-1}{2n}I_{n-1}.$$
$$\implies I_n=\frac{1.3.5.7...(2n-1).(2n-3)}{2.4.6.8...(2n-2).2n}\pi =\frac{(2n-1)!!}{2n!!}\pi$$
