What is a good bound for $\int^{2\pi}_{0}\cos^k(x)dx$? The integral $$\int^{2\pi}_{0}\cos^k(x)dx$$comes up in one of Rudin's problems. He asked me to do a detailed approximation of it. I am wondering how to do it. This feels really elementary, but I need a lower bound that works for any $k$. Integrating by parts would not really help though the function is explicitly integrable. 
 A: If $k$ is odd, then the integral is zero. If $k$ is even, note that $$\int_0^{2\pi} \cos^{2m}(x) dx = 4 \int_0^{\pi/2} \cos^{2m}(x) dx$$
Let $I_m = \displaystyle \int_0^{\pi/2} \cos^{2m}(x) dx$
Then
\begin{align}
I & = \int_0^{\pi/2} \cos^{2m-1}(x) d(\sin(x))\\
& = \left. \cos^{2m-1}(x) \sin(x)\right \vert_0^{\pi/2} + \int_0^{\pi/2}\sin(x) (2m-1) \cos^{2m-2}(x) \sin(x) dx\\
& = (2m-1) \int_0^{\pi/2} \cos^{2m-2}(x) \sin^2(x) dx\\
& = (2m-1) \int_0^{\pi/2} \cos^{2m-2}(x) (1-\cos^2(x)) dx\\
& = (2m-1) \int_0^{\pi/2} \cos^{2m-2}(x) dx - (2m-1) \int_0^{\pi/2} \cos^{2m}(x) dx\\
2mI_{2m} & = (2m-1) I_{2m-2}
\end{align}
Hence, we get that
$$I_{2m} = \dfrac{2m-1}{2m} \cdot \dfrac{2m-3}{2m-2} \cdots \dfrac34 \cdot \dfrac12 I_0$$
Now $I_0 = \int_0^{\pi/2} 1 dx = \dfrac{\pi}2$. Hence, we get that
$$I_{2m} = \dfrac{2m-1}{2m} \cdot \dfrac{2m-3}{2m-2} \cdots \dfrac34 \cdot \dfrac12 \dfrac{\pi}2$$
Hence, $$\int_0^{2\pi} \cos^{2m}(x) dx = 2 \pi \left(\dfrac{2m-1}{2m} \cdot \dfrac{2m-3}{2m-2} \cdots \dfrac34 \cdot \dfrac12 \right) = 2\pi \dfrac{(2m)!}{4^m \cdot  m! \cdot m!} = \dfrac{2 \pi}{4^m} \dbinom{2m}m$$
By Stirling, we have $$\dbinom{2m}m \sim \dfrac{2^{2m+1}}{\sqrt{4 \pi m}} = \dfrac{4^m}{\sqrt{\pi m}}$$
Hence, we have that $$\int_0^{2\pi} \cos^{2m}(x) dx \sim \dfrac{2 \pi}{\sqrt{\pi m}} = \dfrac{2 \sqrt{\pi}}{\sqrt{m}}$$
