Show that $\int_0^{2\pi} \cos^n(kx) \, dx=\int_0^{2\pi} \cos^n(x) \, dx$. 
Show that $\int_0^{2\pi} \cos^n(kx) \, dx=\int_0^{2\pi} \cos^n(x) \, dx$, with $k\in \mathbb N^*$ 

I would really appreciate some help with this exercise because I really don't know how to begin. What I have been able to pick on so far is that $\cos(x)=\cos(kx)=1$  for $x=0$ and $x=2\pi$, and if we raise $1$ to any power it is still $1$. I don't know how I can use this to prove the equality.
 A: Nothing special about $\cos^n$.
All that is needed is
that it is periodic.
If $f(x)$ 
is a periodic function
of period $T$ then
\begin{align}
\int_0^T f(kx)\,dx
&=\frac1{k}\int_0^{kT} f(y) \, dy
\qquad y = kx, dx = dy/k\\
&=\frac1{k}\sum_{j=0}^{k-1}\int_{jT}^{(j+1)T} f(y) \,dy
\qquad\text{split into } k \text{ parts}\\
&=\frac1{k}\sum_{j=0}^{k-1}\int_{0}^{T} f(y+jT) \,dy
\qquad\text{shift origin to zero}\\
&=\frac1{k}\sum_{j=0}^{k-1}\int_0^T f(y) \,dy
\qquad\text{use the fact that it is periodic}\\
&=\frac1{k} \left (k\int_0^T f(y) \, dy\right)
\qquad k\text{ copies of the same thing}\\
&=\int_0^T f(y) \, dy
\qquad\text{Done!}\\
\end{align}
A: By letting $t=kx$, then $dt=k\,dx$ and
$$\begin{align}
\int_0^{2\pi}{\cos^n(kx)}\,dx
&=\int_0^{2\pi k}{\cos^n(t)}\,\frac{dt}{k}
=\frac{1}{k}\sum_{j=1}^k\int_0^{2\pi}{\cos^n(t+2\pi(j-1))}\,dt\\
&=\frac{1}{k}\sum_{j=1}^k\int_0^{2\pi}{\cos^n(t)}\,dt=\int_0^{2\pi}{\cos^n(t)}\,dt
\end{align}$$
where we used the fact that the cosine has period $T=2\pi$ and therefore $$\cos(t+2\pi(j-1))=\cos(t).$$
Along the same lines, you may generalize the property: if $f$ is any integrable function with period $T$ then
$$\int_0^T f(kx)\,dx=\int_0^{T} f(x)\,dx.$$
A: Using the substitution $u = kx$:
$$\int_0^{2\pi} \cos^n kx \; dx = \int_0^{2k\pi} \frac{1}{k} \cos^n u \; du  $$
Since $\cos^n u$ has period $2\pi$, this last integral is equal to
$$\int_0^{2\pi} \frac{1}{k} \cos^n u \; du  +\int_{2\pi}^{4\pi} \frac{1}{k} \cos^n u \; du  +\cdots \int_{2(k-1)}^{2k\pi} \frac{1}{k} \cos^n u \; du,$$
and each of these integrals is equal to the first one (each one is over one complete period.)  And there are $k$ of them, so you have the above 
$$= k\int_0^{2\pi} \frac{1}{k} \cos^n u \; du = \int_0^{2\pi} \cos^n x \; dx$$
which is your other integral once you change $u$ to $x$.
