Using Wallis' formula to compute integral of powers of cos and sine I've recently discovered Wallis' formula to compute powers of cos and sine from $[0,\pi/2]$, However what If I have a function like $\cos^m (x)\sin^n(x)$ where both $m$ and $n$ are even, this function is even, so it must be symmetric to some axis.
If I want to compute the integral of this function but from $[0,k\pi]$. Can I use parity of the function to integrate from $0$ to $\pi/2$ and then use Wallis formula? 
Also I've noticed that for $\cos^m (x)\sin^n(x)$ if one of $m$ or $n$ is odd then the integral on $[0,k\pi]$ is $0$, why is that? 
 A: I have not studied Wallis' integrals in the past, but I will give a try to answer this.
About the first part, a way to solve it in terms of Wallis' integrals is to rewrite either $\sin^n x$ or $\cos ^m x$ using Newton's binomial theorem. $$I(n,m)=\int_0^\frac{\pi}{2}\sin^n x\cos^m x dx=\int_0^\frac{\pi}{2}\sin^n x(1-\sin^2 x)^m dx$$
$$(1-\sin^2 x)^m=\sum_{k=0}^n \binom{m}{k}(-1)^k\sin^{2k}x$$
$$\Rightarrow I(n,m)=\sum_{k=0}^m \binom{m}{k}(-1)^k\int_0^\frac{\pi}{2}\sin^{n+2k}xdx=\sum_{k=0}^m \binom{m}{k}(-1)^k W_{n+2k}$$
I doubt that this is useful at all for large numbers, after all it would have been mentioned somewhere a solution for $I(n,m)$ in terms of Wallis' integral since this is quite a known integral, unfortunately there isn't. An alternative here is to use Beta function:
$$B(a,b)=2\int_0^\frac{\pi}{2} \sin^{2a-1}x\cos^{2b-1}xdx$$
If we set here $2a-1=n\,$ and $2b-1=\,$ we get that:
$$I(n,m)=\int_0^\frac{\pi}{2}\sin^n x \cos^m x dx=\frac{1}{2} B\left(\frac{n+1}{2},\frac{m+1}{2}\right)$$
About your second question. Yes, you can use the periodicity of the sine and cosine function to conclude that: $$\int_0^{k \pi}\sin^{2n}x \cos^{2m}xdx=k\int_0^{\pi} \sin^{2n}x \cos^{2m} xdx=2k\int_0^{\frac{\pi}{2}} \sin^{2n}x \cos^{2m} xdx$$
Although it's simpler if one looks at the graph we can prove this directly, starting by splitting the interval into $k$ subintervals, namely:
$$I(2n,2m)=\int_0^\pi \sin^{2n}x \cos^{2m} xdx+\int_{\pi}^{2\pi}\sin^{2n}x \cos^{2m} xdx+\dots+\int_{(k-1)\pi}^{k\pi}\sin^{2n}x \cos^{2m} xdx$$
$$=\sum_{j=1}^{k}\int_{(j-1)\pi}^{j\pi}\sin^{2n}x \cos^{2m} xdx$$
Letting $\,\displaystyle{x-(j-1)\pi=t\Rightarrow x=t+(j-1)\pi \Rightarrow dx=dt}$ and using:
$$\sin(t+(j-1)\pi))=\sin t \cos((j-1)\pi)=(-1)^{j-1}\sin t$$
$$\cos(t+(j-1)\pi))=\cos t \cos((j-1)\pi))=(-1)^{j-1}\cos t$$
$$I(2n,2m)=\sum_{j=1}^k \int_0^\pi ((-1)^{j-1}\sin t)^{2n}((-1)^{j-1}\cos t)^{2m}dt$$
Well, what happens if we take $(-1)^{j-1}$ to an even power? Of course: $\left((-1)^{j-1}\right)^{2k}=1$. Thus we get $k$ integral and everyone equal to each other.
$$I(2n,2m)=\sum_{j=1}^k \int_0^\pi \sin^{2n}x \cos^{2m} x dx=k\int_0^\pi \sin^{2n}x \cos^{2m} x dx$$ I will let you split the integral as: $\,\displaystyle{\int_0^\frac{\pi}{2}+\int_\frac{\pi}{2}^{\pi}}$ and think of a substitution in order to show that: $$\int_0^\pi \sin^{2n}x \cos^{2m} x dx=2\int_0^\frac{\pi}{2} \sin^{2n}x \cos^{2m} x dx$$
Also about your third question we can use the fact that: $\displaystyle{\int_a^b f(x)dx= \int_a^b f(a+b-x)dx}$
$$\int_0^{k\pi} \sin^n x \cos^m xdx=\int_0^{k\pi} (\sin(k\pi -x))^n(\cos(k\pi-x))^mdx$$
$$\sin(k\pi -x )=-\sin x\cos(k\pi)=(-1)^{k+1}\sin x$$
$$\cos(k\pi -x )=\cos x\cos(k\pi)=(-1)^k\cos x$$
Thus if $n$ is odd and $m$ is even we get the integral to be:
$$J=\int_0^{k\pi} (-1)^{(k+1)(2n+1)}\sin^{2n+1}x (-1)^{k(2m)}\cos^{2m}x dx=-\int_0^{k\pi}\sin^{2n+1}x\cos^{2m} xdx$$
And well since: $\displaystyle{J=-J \Rightarrow 2J=0\Rightarrow J=0}$. By the same argument we can show that if $n$ and $m$ are not both even then the integral vanishes.
A: @Zacky showed you that identity with the Beta function. Here's how to derive it.
Consider the integral 
$$I(a,b)=\int_0^{\pi/2}\sin(x)^a\cos(x)^b\mathrm dx$$
Straight away, we preform the substitution $t=\sin(x)^2$:
$$
\begin{align}
I(a,b)=&\frac12\int_0^1t^{a/2}(1-t)^{b/2}t^{-1/2}(1-t)^{-1/2}\mathrm dt\\
=&\frac12\int_0^1t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}\mathrm dt\\
=&\frac12\int_0^1t^{\frac{a+1}2-1}(1-t)^{\frac{b+1}2-1}\mathrm dt\\
\end{align}
$$
Next we recall the definition of the Beta function:
$$B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}\mathrm dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
Hence we have 
$$I(a,b)=\frac12B\bigg(\frac{a+1}2,\frac{b+1}2\bigg)$$
